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All Solutions of Type: Mechanics
 0 Click here to jump to the problem! GR8677 #1 Mechanics$\Rightarrow$}Newtonian This problem can be translated into an equation: $m\ddot{x}=-k\dot{x}-mg. $ Note that the sign is compensated for. $\dot{x}$ is positive going up (so that $-k\dot{x}$ acts in the same direction as gravity), and it is negative going down (acting in the opposite direction of gravity). \par (A) $\ddot{x}\neq g \forall t$ because $\ddot{x}=\frac{1}{m}\left(-k\dot{x}-mg\right)$ (B) True. $\dot{x}=0$ at the top. Thus, plug that into the equation in described in (A) and get $\ddot{x}=-g$ (C) $\ddot{x}=\frac{1}{m}\left(-k\dot{x}-mg\right)$ could be greater than $g$, especially when $\dot{x}$ is positive in sign. (D) The wind friction ($-k\dot{x}$) term would change sign when going downwards. (See the Note above.) Thus, the equation of motion presented above would be different in sign going down and up. (Also, one could argue that after the force of wind friction acts, to conserve energy, kinetic energy must lessen in the final energy sum. Thus, the final velocity can't be the same as $v_0$. Or, intuitively, one could see friction as a symmetry breaker, viz., the ideal parabolic path is disturbed in such a way that the end velocity changed.) (E) Since the force other than gravity is a frictional force, which, by definition, slows down the object, the final velocity has to be less than $v_0$. Click here to jump to the problem!
 1 Click here to jump to the problem! GR8677 #2 Mechanics$\Rightarrow$}Vector While there may be a more quantitative solution, the simplest solution is qualitative, based on elementary vector addition and knowledge of the force center. The problem states that the object orbits the Earth in a perfect circle, initially. This means that the initial velocity ($\vec{v_0}$) is perpendicular to the vector pointing to the earth center ($\vec{n}$), i.e., it's tangent to the circular path. This is the condition for uniform circular motion (the centripetal acceleration is due to the Gravitational Law). After firing a missile straight to Earth center, its velocity gains an extra normal component ($-\vec{v_n}$), equal and opposite to the velocity of the missile fired to Earth. Thus, its trajectory would deviate from the circular trajectory. Because the only source of acceleration comes from the Earth center, $-\vec{v_n}$, which is parallel to the centripetal acceleration provided by the Earth, will eventually go to 0. Recall that acceleration does not effect velocity components in the perpendicular direction (to wit: a projectile fired on Earth has the same constant $v_x$, but its $v_y$ changes). There will thus always be a (nearly constant) tangential velocity, even at the perigees. However, $-\vec{v_n}$ will go to 0 at the perigees. The tangential velocity will remain more-or-less constant, so that instead of spiraling inwards, the path becomes an ellipse, as $-\vec{v_n}$ is restored at the apogees and zero'ed at the perigees. (In a more down-to-earth form, this problem is essentially a projectile firing question with no numerical work involved.) Click here to jump to the problem!
 2 Click here to jump to the problem! GR8677 #5 Mechanics$\Rightarrow$}Conservation This is a three step problem involving conservation of energy in steps one and three and conservation of momentum in step two. 1. $Mgh_0=\frac{1}{2}Mv_{1}^{2}$... Conservation of energy determines the velocity of putty A, $v_1$, (of mass $M$) at the bottom of its trajectory, right before it intersects B. 2. $Mv_1 = (M+3M)v_2=4Mv_2$... Since the putty thingies stick together, conservation of momentum is easy. Solve for $v_2$. 3. $\frac{1}{2}4Mv_{2}^{2}=4Mgh_3$... Conservation of energy, again. Solve for $h_3$. And voila, plug in the relevant quantities to find that the answer is (A) $\frac{1}{16} h_0$. Click here to jump to the problem!
 3 Click here to jump to the problem! GR8677 #6 Mechanics$\Rightarrow$}Vectors Since there is only one force acting, i.e., the gravitational force, one can find the tangential acceleration by projecting $\vec{g}$ in the tangential direction. Equivalently, one dots gravity with the tangential unit vector, $\vec{v_t}=\vec{g} \cdot \hat{t}$. There's a long way to do this, wherein one writes out the full Gibbsean vector formalism, and then there's a short and elegant way. (The elegant solution is due to Teodora Popa.) The problem gives $f(x) = y = x^2/4$. Thus, $df/dx = dy/dx = x/2 = \tan\alpha$, where in the last step, one notes that the ratio $dy/dx$ forms the tangent of the indicated angle. One recalls the Pythagorean identity $\sin^2\alpha + \cos^2\alpha = 1$, and the definition of $\tan\alpha$ in terms of $\sin\alpha$ and $\cos\alpha$. Thus, one gets $x/2=\tan\alpha = \frac{\sin\alpha}{\cos\alpha}=\frac{\sin\alpha}{\sqrt{1-\sin^2\alpha}}$. Square both sides to get $x^2/4 = \tan^2\alpha = \frac{\sin^2\alpha}{\cos^2\alpha}=\frac{\sin^2\alpha}{1-\sin^2\alpha}$. Solve $x^2/4 = \frac{\sin^2\alpha}{1-\sin^2\alpha}$ to get $\sin^2 \alpha = \frac{x^2/4}{1+x^2/4} = \frac{x^2}{4+x^2}$. The angle between the vectors $\vec{g}$ and $\hat{t}$ is $\pi/2-\alpha$, and thus the tangential acceleration is $g \sin \alpha = \frac{g x}{\sqrt{x^2+4}}$. Beautiful problem. Click here to jump to the problem!
 4 Click here to jump to the problem! GR8677 #7 Mechanics$\Rightarrow$}Statics Straight-forward Newtonian statics: $F_x=0=F-T\sin(\theta)\Rightarrow T\sin(\theta)=F$ $F_y=0=T\cos(\theta)-mg\Rightarrow T\cos(\theta)=mg$ Divide the two equations above, cancel T's, and get: \par $\tan(\theta)=\frac{F}{mg}=\frac{10}{2*10}=.5$. Choice A is right. Click here to jump to the problem!
 5 Click here to jump to the problem! GR8677 #8 Mechanics$\Rightarrow$}Conservation Straight-forward conservation of energy problem: $E_0=\frac{1}{2}m v^2 = E_f=Fl, $ where $l=0.025 m =\frac{25}{1000}$, $m=5kg$, and $v=10 m/s$. Plug everything in, and solve for $F$, to get choice D, $10,000 N$. Click here to jump to the problem!
 6 Click here to jump to the problem! GR8677 #34 Mechanics$\Rightarrow$}Potential Energy Recall the lovely relation, $F_x=-\frac{\partial U}{\partial x} \Leftrightarrow \vec{F}=-\nabla U.$ $U=kx^4\Rightarrow F=-4kx^3$ Click here to jump to the problem!
 7 Click here to jump to the problem! GR8677 #35 Mechanics$\Rightarrow$}Hamiltonian Recall the lovely relation, $H=T+V,$ where $T$ is the kinetic energy and $V$ is the potential energy. (And, while one is at this, one should also recall that the Lagrangian is $L=T-V$.) The potential energy is given to be $kx^4$. The kinetic energy $\frac{1}{2}mv^2$ can be written in the form of $\frac{p^2}{2m}$. Thus, choice (A) is the right answer. Click here to jump to the problem!
 8 Click here to jump to the problem! GR8677 #36 Mechanics$\Rightarrow$}Lagrangian Recall Hamilton's Principle (of least action), $\int_{t_1}^{t_2} L dt, $ where $L=T-V$ is the Lagrangian and $T$ is the kinetic energy and $V$ the potential energy. The potential energy is given in the problem. The choice is obviously (A). Click here to jump to the problem!
 9 Click here to jump to the problem! GR8677 #37 Mechanics$\Rightarrow$}Statics Sum over each component. Note that the horizontal direction has a net force proportional to the centripetal acceleration $a=v^2/r = \omega^2 r$, where $v=\omega r$. Note that $T$ is the tension. $\begin{eqnarray} \sum F_x=T \sin(\theta/2)&=&mv^2/r=m\omega^2r\\ \sum F_y=T \cos(\theta/2)-mg&=&0= \end{eqnarray}$ Solve for T above to get, $\begin{eqnarray} \sum T \sin(\theta/2)&=&m\omega^2r\\ \sum T \cos(\theta/2)&=&mg \end{eqnarray}$ Find the magnitude of T and use the Pythagorean theorem, $||T||=\sqrt{T^2 \cos^2(\theta/2)+T^2 \sin^2(\theta/2)}=m\left(g^2 + \omega^4 r^2 \right)^{1/2},$ and thus (E) is the right answer. (The above should be fairly obvious, but if one is totally clueless, then one can eliminate choice (D) from noting units. The angular velocity has units of $1/s$ but $g^2$ has time units proportional to $1/s^4$.) Click here to jump to the problem!
 10 Click here to jump to the problem! GR8677 #42 Mechanics$\Rightarrow$}Scattering Cross Section The target nuclei per $cm^3$ is $\rho=N/V=1E20 nuclei/cm^3$. The scatterer thickness is $t=1E-1cm$. Passing through the scatterer, only $P=1E-6$ particles are scattered. (This is the probability of scattering.) One can obtain to formula for the scattering cross section from ordinary dimensional analysis spiced up with some common sense (or recalling its definition back in Marion and Thornton's Dynamics book). It is: $A = P\frac{V}{Nl}=\frac{P}{\rho l}=\frac{1E-6}{1E20 \times 1E-1}=10^{-6-20+1}=10^{-25}, $ which is choice (C). Click here to jump to the problem!
 11 Click here to jump to the problem! GR8677 #43 Mechanics$\Rightarrow$}Normal Modes Because there are two degrees of freedom in this problem, there are two normal mode frequencies. Because there is no external torque acting on the system, the center of mass of the system stays the same through time. From common sense, one deduces that $\omega_1$ has to do with the outer masses moving perfectly out of phase, i.e., masses A and C moving either towards the left and right (away from each other), respectively, or right and left (towards each other), respectively---and B being perfectly stationary, thereby conserving" the center of mass. The other angular frequency, $\omega_2$, has to do with either masses A and C moving in phase and mass B out of phase. $\omega_1$ is actually equivalent to having a single mass on a string, since because the middle mass doesn't move, it acts as a sort of support for the spring. $\omega_1 = \sqrt{k/m}$, which would correspond to choice (B). (Of course, one should recall the obvious, that $\omega = 2\pi f$.) (Incidentally, one can derive $\omega_2$ without having to resort to the formalism of matrix mechanics: Since the center of mass remains 0, one has $R = \frac{mx-2mX+mx}{m+2m+m}$. Solving, one gets $X=x$. The displacement of the middle mass, mass B, is thus $x+X=2x$, while the displacements of the smaller masses, masses A and C, are both $x$. The displacement of each spring is $2x$. Potential energy is thus $U=2\frac{1}{2}k(2x)^2=\frac{1}{2}8kx^2$. The kinetic energy is $T=2\frac{1}{2}m(\dot{x})^2+\frac{1}{2}(2m)(\dot{x})^2=\frac{1}{2}4m(\dot{x})^2$. The normal mode frequency is deduced by $\omega_2=\sqrt{\frac{k_{eff}}{m_{eff}}}=\sqrt{\frac{8k}{4m}}=\sqrt{\frac{2k}{m}}$) Click here to jump to the problem!
 12 Click here to jump to the problem! GR8677 #44 Mechanics$\Rightarrow$}Conservation of Momentum One could use energy, but then one would have to take into account the inertia. Momentum might be easier, $(p_i=mv) = (p_f=MV)\Rightarrow V=\frac{m}{M}v,$ where the final momentum takes into account the fact that the final velocity of the particle is at rest (0). And, so it is (A)! Click here to jump to the problem!
 13 Click here to jump to the problem! GR8677 #60 Mechanics$\Rightarrow$}Simple Harmonic Motion Recall $F=-\frac{\partial V}{\partial x}=-2bx$. Simple harmonic motion has the simple form, $\ddot{x}\propto x$. Thus $F=m\ddot{x}=-2bx \Rightarrow \ddot{x}=-\omega^2 x$, where $\omega^2=2b/m$ is the frequency squared. Thus, simple harmonic motion occurs with a frequency determined by both $b$ and $m$. This is choice (C). Click here to jump to the problem!
 14 Click here to jump to the problem! GR8677 #61 Mechanics$\Rightarrow$}Conservation of Momentum One can easily derive the formula for rocket motion, $\begin{eqnarray} p_{initial}&=&p_{final}\\ mv &=& (m-dm')(v+dv)+(dm')(v-V)\\ mv &=& mv + mdv - dm'v - dm'dv + dm'v - dm'V\\ mdV &\approx& dm'V\\ mdV &=& -dm V, \end{eqnarray}$ where $m$ is the mass of the rocket and $m'$ is the mass of the fuel rejected. $v$ is the initial velocity of the rocket. $V$ is the velocity of the exhaust. The final line comes about from realizing that $dm=-dm'$. The derivation starts with the assumption that the final mass of the rocket is its original mass minus $dm'$, the eject mass, and that its final velocity is a tiny bit faster than it was before $dv$. The exhaust mass' velocity relative to the rocket is $V$. (Higher order terms such as $dxdy$ have been thrown out to arrive at the final differential equation.) $V$ is equivalent to the velocity of the exhaust mass relative to the rocket, the inertial reference system. Thus $u=V$ and the answer is choice (E). (Note that firing the exhaust backwards generates a forward thrust for the rocket---motion in space is quintessentially dependent on the phenomenon of farting. So, if one is well-equipped and wanna get going in space, just let out a bit of gas, and one's good to go.) Click here to jump to the problem!
 15 Click here to jump to the problem! GR8677 #62 Mechanics$\Rightarrow$}Separable Differential Equations Solve $m\frac{dv}{dt}=-u\frac{dm}{dt}$. Separate variables to get \par $\int u\frac{dm}{m}=\int -dv\Rightarrow u \ln (m_0/m)=v(t)$. None of the answer choices such a ln relation for $v$, and thus the answer is (E), none of the above. Click here to jump to the problem!
 16 Click here to jump to the problem! GR8677 #75 Mechanics$\Rightarrow$}Kepler's Third Law One recalls (or should memorize now) the famous slogan: The square of the period is proportional to the cube of the distance. In equation form, $T^2 = k d^3$ One does not need to know the proportionality factor to solve this problem. In fact, the problem gives $T=80$ minutes at $d=R_e$. Thus, $80^2=k R_e^3 \Rightarrow k = 80^2minutes/R_e^3$. There are $24 \times 60$ minutes in 24 hours. Plug that into the equation (and the relation for $k$ as determined above) to get, $(24 \times 60)^2 = 80^2/R_e^3 d^3 \Rightarrow d=(24 \times 6/8)^{2/3} R_e$ No calculators allowed, so $(18^2)^{1/3}$ is about $400^{1/3}$, which is closest to $7$, as in choice (B). Click here to jump to the problem!
 17 Click here to jump to the problem! GR8677 #76 Mechanics$\Rightarrow$}Conservation of Energy Recall the formula for inertia $I=\int dm r^2$. A hoop has constant $r=R$, thus the integral is trivial and $I_{hoop}=MR^2$. One can break up the kinetic energy into the pure rolling (about center of mass) and pure translation bit, $\begin{eqnarray} Mgh&=&\frac{1}{2}I\omega^2 + \frac{1}{2}Mv^2\\ &=&\frac{1}{2}MR^2\omega^2 + \frac{1}{2}Mv^2\\ Mgh&=&MR^2\omega^2\\ gh&=&R^2\omega^2\\ &\Rightarrow& \omega = \sqrt{\frac{gh}{R^2}} , \end{eqnarray}$ where one recalls that $v=R\omega$. Plug the value of angular velocity $\omega$ into the equation for angular momentum $L=I\omega$, to get $MR^2 \sqrt{\frac{gh}{R^2}}=MR\sqrt{gh}$, as in choice (A). Click here to jump to the problem!
 18 Click here to jump to the problem! GR8677 #78 Mechanics$\Rightarrow$}Conservation of energy There isn't much to say about this problem other than the fact that it's a conceptual conservation of energy problem. Particle 1 moves at velocity $v$ towards particle 2, initially at rest. EM-potential energy increases as they get closer and closer together---but, energy should still be conserved since no energy is radiated. The potential energy increase comes from the initial kinetic energy of particle 1. Take choice (C). (The other choices are all either too weird or just plain bogus.) Click here to jump to the problem!
 19 Click here to jump to the problem! GR8677 #93 Mechanics$\Rightarrow$}Power Recall the following basic formulas, $P=IV=Fv$, where $P$ is power, $I$ is current, $V$ is voltage, $F$ is force, and $v$ is velocity. $e=\frac{Work_{done}}{Work_{input}}$, where $e$ is the efficiency, which relates work (and thus power). The problem gives $e=1$, $F=\mu N=100\mu$, $v=10$, $I=9$, $V=120$---where all units are SI. Thus $e=1 \Rightarrow P_{done}=P_{input} \Rightarrow \mu N v = IV$. Solve for $\mu=\frac{IV}{Nv}=\frac{9*120}{100*10}=108/100\approx 1.1$, as in choice (D). Click here to jump to the problem!
 20 Click here to jump to the problem! GR8677 #97 Mechanics$\Rightarrow$}Conservation of Momentum The rotational part of the angular momentum is $L=I\vec{\omega_0}$. The translational part of the angular momentum is $L_{v_0}=m \vec{r} \times \vec{v_0}$. (Note that, according to the diagram, this cross-product points in the other direction to the angular velocity.) Initially, the angular momentum about the point P is $L=\frac{1}{2}MR^2 \omega_0 - L_{v_0} = 0$, since $v_0=\frac{1}{2}R^2 \omega_0^2$ and $\vec{L_{v_0}}=\vec{r}\times\vec{p}=-RMv_0=-\frac{1}{2}M\omega_0 R^2$. QED.Click here to jump to the problem!
 21 Click here to jump to the problem! GR8677 #5 Mechanics$\Rightarrow$}Centripetal Force $\vec{F}_air$ acts in the direction as shown and the centripetal acceleration acts in the direction of $\vec{F}_A$. Centripetal acceleration is a net force, however, and thus, $\begin{eqnarray} \rightarrow_+ \sum F_x = 0 &=& -F_{air} +f_x \\ \uparrow_+ \sum F_y = -mv^2/r &=& f_y \end{eqnarray}$ $f_x$ is in the positive direction and $f_y$ is in the negative direction. Thus the force of the road is $\vec{F}_B$. Click here to jump to the problem!
 22 Click here to jump to the problem! GR8677 #6 Mechanics$\Rightarrow$}Inclined Plane Set up the usual coordinate system with horizontal axis parallel to incline surface. The equations are, (since the mass slides down at constant speed), $\begin{eqnarray} \sum F_x &=& 0 = f-mg\sin\theta\\ \sum F_y &=& 0 = N-mg\cos\theta \end{eqnarray}$ Friction is given by $f=\mu N =\mu mg\cos\theta$, where the normal force $N$ is determined from the $F_y$ equation. For constant velocity one also has, $f=mg\sin\theta=\mu mg\cos\theta\Rightarrow \mu = \tan \theta$ To find the work done by friction, one calculates $W=f L$, where $L\sin\theta = h$. Thus $W = \tan\theta mg \cos\theta \frac{h}{\sin\theta}=mgh$, as in the almost-too-trivial, but right, choice (B). Click here to jump to the problem!
 23 Click here to jump to the problem! GR8677 #7 Mechanics$\Rightarrow$}Elastic Collisions One determines the velocity of impact of the ball from conservation of energy, $mgh = \frac{1}{2}mv_0^2 \Rightarrow v_0^2 = 2 gh$ Conservation of momentum gives, $v_0=-v_1 + 2v_2$ Conservation of kinetic energy gives, $mv_0^2 = mv_1^2+2v_2^2$ Plug in the momentum and kinetic energy conservation equations to solve for $v_1$ and $v_2$ in terms of $v_0$ to get $\begin{eqnarray} v_1&=&-v_0/3\\ v_2&=2v_0/3& \end{eqnarray}$ Write yet another conservation of energy equation for the final energy, $\frac{1}{2}mv_1^2 +\frac{1}{2}2mv_2^2 = mgh' +\frac{1}{2}2m v_2^2,$ where the condition that the mass $2m$ slides on a frictionless plane is used. Thus, $\frac{1}{2}mv_1^2 = \frac{v0^2}{18} = mgh'\Rightarrow h'=\frac{h}{9}$, where the previous result $v_1 = -v_0/3$ and $v_0^2 =2gh$ is used. Click here to jump to the problem!
 24 Click here to jump to the problem! GR8677 #8 Mechanics$\Rightarrow$}Damped Oscillations One should remember that damped oscillations have decreasing amplitude according to an exponential envelope. As the amplitude shrinks, the period increases. The additional force instated in the problem is equivalent to damping, and thus the period increases, as in choice (A). Click here to jump to the problem!
 25 Click here to jump to the problem! GR8677 #20 Mechanics$\Rightarrow$}Conservation of Momentum The Helium atom ($m$) makes an elastic collision, and thus the kinetic energy before and after is conserved. $\frac{1}{2}mv^2=\frac{1}{2}m(0.6v)^2+\frac{1}{2}MV'^2 \Rightarrow 0.64mv^2 = MV'^2$ Conservation of momentum requires that, $mv = -.6mv +MV' \Rightarrow V' = 1.6mv/M$ From kinetic energy conservation, $0.64mv^2 = MV'^2\Rightarrow 0.64mv^2 = (1.6mv)^2/M \Rightarrow 0.64=1.6^2m/M\Rightarrow M=1.6^2m/0.64=4m$, but since $m=4u$, $M=16u$, for $O_2$, as in choice (D). Click here to jump to the problem!
 26 Click here to jump to the problem! GR8677 #21 Mechanics$\Rightarrow$}Moment of Inertia To solve this problem, one should remember the parallel axis equation to calculate the moment of inertia about one end of the hoop: $I = I_{cm} + md^2 = mR^2+md^2=2mR^2$, where $d$ is the distance from the pivot point to the center of mass, which in this problem, is just equal to $R$. (In the last equality, note that the moment of inertia of a hoop of radius R and mass m about its center of mass is just $I_{cm}=mR^2$.) The problem gives the period of a physical pendulum as $T=2\pi \sqrt{I/(mgd)}$. Thus, plugging in the above result for the moment of inertia, one has, $T=2\pi \sqrt{2mR^2/(mgR)}=2\pi \sqrt{2R/(g)} \approx 2*3\sqrt{2*0.2/(10)}= 12/10=1.2s$, which is closest to choice (C). (Since $\pi$ was rounded to 3, the period should be slightly longer than 1.2s.) Click here to jump to the problem!
 27 Click here to jump to the problem! GR8677 #22 Mechanics$\Rightarrow$}Geometry The harder part of this problem involves determining the radius of Mars. It's an approximate geometry problem. The problem gives a vertical drop of 2m for every 2600m tangent to the surface. The tangent to the surface is approximately one leg of a triangle whose hypotenuse is the radius of Mars, since the radius is much larger than the tangent distance. The other leg of the right triangle is just $r-2$, where $r$ is the radius of Mars. In equation form, what was just said becomes $(r-2)^2+3600^2=r^2$. The square terms cancel out, and dropping out the 4, one has $r\approx 3600^2/2 \approx 8E6 m$. (The above deduction was due to Ayanangsha Sen.) The easier part comes in the final half of the problem: applying the centripetal force to the force of gravity. $mv^2/r = mg \Rightarrow v=\sqrt{2 g r}\approx \sqrt{20 * 8E6}=\sqrt{16E7}\approx 4000 m/s$, which is closest to $3.6km/s$, as in choice (C). Click here to jump to the problem!
 28 Click here to jump to the problem! GR8677 #23 Mechanics$\Rightarrow$}Stability of Orbits The gravitational force suspect to a bit of perturbation is given as $\vec{F}_{12} = \hat{r}_{12} Gm_1 m_2/r_{12}^{2+\epsilon$. One can narrow down most choices by recalling some basic facts from central force theory: (A) No mention is made of frictional effects, and thus energy should be conserved. (B) Angular momentum is always conserved since the net torque is 0 (to wit: the force and moment arm are parallel). (C) This is just Kepler's Third Law applied to this force. (Recall the following bromide: The square of the period is equal to the cube of the radius---for the inverse square law force. For a perturbed force, the bromide becomes: The square of the period is equal to the $3+\epsilon$ power of the radius.) (D) Recall Bertrand's Theorem from Goldstein. Stable non-circular orbits can only occur for the simple harmonic potential and the inverse-square law force. This is of neither form, and thus this choice is FALSE. (E) Circular orbits exist for basically all potentials. A stationary orbit exists if and only if the following conditions are satisfied: $V'=0 \; \;V''>0$. Recall that the potential is related to the force by $-V'=F\Rightarrow V=-\int F dx$. Use $V\propto 1/r^n$, and recalling the extra term added to the effective potential to be $L^2/(2mr^2)$, one chunks out the derivatives to get the condition that $n<2$, as a potential exponent, ($n<3$, as a force exponent) for stable orbit. One can remember this result or re-derive it whenever necessary. For $n < 3$, (the power exponent of the force equation), a stable circular orbit exists. Since $\epsilon$ is presumably less than 1, the planet does, indeed, move in a stationary circular orbit about the sun. Click here to jump to the problem!
 29 Click here to jump to the problem! GR8677 #31 Mechanics$\Rightarrow$}Frictional Force (A) A falling object experiencing friction falls faster and faster until it reaches a terminal speed. Its kinetic energy increases proportional to the square of the velocity and approaches a asymptotic value. (B) The kinetic energy increases to a maximum, but it does not decrease to 0. See (A). (C) The maximal speed is the terminal speed. (D) One has the equation $m\ddot{y}+b\dot{y}+mg=0\Rightarrow m\dot{v}+bv+mg=0$. Without having to solve for $v$, one can tell by inspection that $v(t)$ will depend on both $b$ and $m$. (E) See (D). This is the remaining choice, and it's right. Click here to jump to the problem!
 30 Click here to jump to the problem! GR8677 #32 Mechanics$\Rightarrow$}Moment of Inertia The inertia through the point A is $I_A=3mr^2$. From geometry, one deduces that the distance between each mass and the centerpoint A is $r\cos(30^\circ)=l/2\Rightarrow r=l/\sqrt{3}$. T he moment of inertia about A is thus $I_A=ml^2$ The inertia about point B can be obtained from the parallel axis theorem ($I_{displaced}=I_{cm}+\sum_i m_i d^2$, where d is the displaced distance from the center of mass). Because $d=l/\sqrt{3}$, one has $I_B = I_A + 3md^2 =2 I_A$. Since the angular velocity is the same for both kinetic energies, recalling the relation for kinetic energy $K_i=I_i \omega_i^2$, one has $K_B/K_A=I_B/I_A=2$, as in choice (B). Click here to jump to the problem!
 31 Click here to jump to the problem! GR8677 #44 Mechanics$\Rightarrow$}Chain Rule Recall that $a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$. $\frac{dv}{dx}=-n\beta x^{-n-1}\Rightarrow v\frac{dv}{dx}=-n\beta^2 x^{-2n-1}$, as in choice (A). Click here to jump to the problem!
 32 Click here to jump to the problem! GR8677 #65 Mechanics$\Rightarrow$}Conservation Laws From conservation of momentum, one has $mv_0=Mv \Rightarrow v_0 = Mv/m$. The man does work on both himself and the boat. Thus, the work-kinetic energy theorem has $W=\Delta K = 1/2 m v_0^2 + 1/2 M v^2 = 1/2( M^2/m +M )v^2, as in choice (D)$ Click here to jump to the problem!
 33 Click here to jump to the problem! GR8677 #66 Mechanics$\Rightarrow$}Effective Potential One can solve this problem by remembering the effective potential curve $V_{eff}(r) = V(r)+L^2/(2mr^2)$. For the gravitational potential, one has $V(r) \propto - 1/r$. The total energy of the spaceship is $E_s = 1/2 m(1.5v_J)^2+V_{eff}$, while the total energy of Jupiter is $E_J = 1/2 m(v_J)^2 + V_{eff}$ (A) A spiral orbit occurs when $E, which corresponds to $v_s<. (B) A circular orbit occurs only when $E = V_{min}$. Since the energy of Jupiter is greater than that of the spaceship--and (see below) since Jupiter itself has $E>V_{min}$, the spaceship must have $E>V_{min}$. (C) An ellipse occurs for $V_{min}. Planets orbit in ellipses. However, since the speed of the ship is greater than Jupiter's orbit speed by a good bit, one assumes its total energy is $E>0$. (D) A parabolic orbit occurs for $E=0$. The condition is much too stringent. (E) A hyperbolic orbit occurs for $E>0$. See (C). Since $E_s>0>E_J$, this is it. Click here to jump to the problem!
 34 Click here to jump to the problem! GR8677 #68 Mechanics$\Rightarrow$}Lagrangians The potential energy of the mass is obviously $U=mgs\cos\theta$, and thus one eliminates choices (B) and (C). (To wit: $L=T-U$, (C) has the wrong sign). The translational part of the kinetic energy is easily just $1/2 m \dot{s}^2$. The rotational part requires the calculation of the moment of inertia for a point particle, which is just $I=mr^2$, where $r=s\sin\theta$, in this case. Thus, the rotational kinetic energy is $1/2 m (s\sin\theta)^2 \omega^2$. The only choice that has the right rotational kinetic energy term is choice (E). Click here to jump to the problem!
 35 Click here to jump to the problem! GR8677 #80 Mechanics$\Rightarrow$}Wave Phenomena There's a long way to solve this problem and then a short. One looks at the choices to find the one that first the physical deduction: when $\mu_l=\mu_r$, the whole incident wave should be transmitted, with 0 reflection. Moreover, in the limit of $\mu_r>>\mu_l$ there should be 0 transmission. Choice (C) is the only one that fits this condition, leading to a ratio of 1 for $\mu_l=\mu_r$. One can also calculate the exact form of the transmission coefficient for this multi-density string. Take the following, $\begin{eqnarray} \psi_i = Re \left(e^{-i(k_l x-\omega t)} \right)\\ \psi_r = \left( Re R e^{-i(-k_l x-\omega t)} \right)\\ \psi_t = \left( Re T e^{-i(k_r x-\omega t)} \right) \end{eqnarray}$ At the boundary between different density parts, one applies continuity $\psi_i(x=0)+\psi_r(x=0)=\psi_t(x=0)$ to get 1+R=T. One applies $m \ddot{\psi} = 0 = \frac{\partial \psi}{\partial x}(x<0) - \frac{\partial \psi}{\partial x}(x>0)$, where $m=0$ since there is no point particle situated at the origin, to obtain $ik_l(1-R)=ik_r T$. Recalling the nifty relation $\omega = ck$ and $c=\sqrt{F/\mu}$, one solves for T to get $T=\frac{2k_l}{k_r+k+l}=\frac{\sqrt{\mu_l/\mu_r}}{1+\sqrt{\mu_l/\mu_r}}$, as in choice (C). Click here to jump to the problem!
 36 Click here to jump to the problem! GR8677 #83 Mechanics$\Rightarrow$}Rippled Surface The simple intuitive way to solve this is to note that for $d \rightarrow \infty$, $v \rightarrow 0$, since one gets an infinitely steep (vertical line) hill, and the only way for the particle to stay on the surface (i.e., not accelerate on it) at the vertical drop is if its velocity is 0. The onyl choice with d on the denominator is choice (D). The more rigorous solution is due to Sara Salha. Equating centripetal force with gravity at the top of the hill, one has $mv^2/r=mg \Rightarrow v=\sqrt{mgr}$. The non-trivial bit comes from calculating the radius. Recall the radius of curvature from calculus $1/r = \kappa = \frac{|\ddot{x}\dot{y}-\ddot{y}\dot{x}|}{(\dot{x}^2+\dot{y}^2)^{3/2}}$. Defining a parameter t as the independent variable, and defining $x=t$, $y=d\cos(kt)$, one finds that $1/4 = \frac{k^2d \cos(kt)}{1+(kd)^2\sin^2(kt)}$. Evaluate it at $t=0$ to find $1/r(0)=k^2d \Rightarrow r=1/(k^2d)$, the radius of curvature at the top of the hill. Plug that into the equation for velocity above to get $v=\sqrt{\frac{mg}{kd^2}}$, as in choice (D). Click here to jump to the problem!
 37 Click here to jump to the problem! GR8677 #84 Mechanics$\Rightarrow$}Normal Mode One can work through the formalism of the usual normal mode analysis or learn how to deal with normal mode frequencies the easy way: The highest normal mode frequency is due to the two masses oscillating out of phase. The $\omega^2$ contribution from the pendulum is just $g/l$. The $\omega^2$ contribution from each mass due to the spring is $k_i/m_i$. This is choice (D). (If one had to guess, one can immediately eliminate choice (A), since that is the lowest normal mode frequency. In normal modes, there's always an in-phase frequency, which tends to be the lowest frequency.) Click here to jump to the problem!
 38 Click here to jump to the problem! GR8677 #85 Mechanics$\Rightarrow$}Wave Phenomena One can solve this problem without knowing anything about mechanics (but with just the barest idea of wave phenomenon theory). Following the hint, one considers the limiting cases: $M\rightarrow \infty \Rightarrow \mu/M \rightarrow 0$... With an infinite M, the string is basically fixed on the rod, and its wavelength is just $\lambda=L$. One eliminates choice (A) from the fact that $\cos 2\pi = 1 \neq 0$, as $\mu/M$ demands in this regime. $M\rightarrow 0 \Rightarrow \mu/M \rightarrow \infty$... Without the presence of the mass M, one has $\lambda = 4L$. Thus, $2\pi L/\lambda=\pi/2$. Since $\tan x = \sin x/\cos x$ and $\cos \pi/2 = 0$, one finds that choice (B) is the only one that fits this condition. Click here to jump to the problem!
 39 Click here to jump to the problem! GR8677 #92 Mechanics$\Rightarrow$}Potential Given $V(x)=-ax^2+bx^4$, one can find the minimum by taking the first derivative (second derivative test indicates concave up), $V'(x) =\left( -2ax+4bx^3\right)_{x_0}=0 \Rightarrow x_0=\sqrt{\frac{a}{2b}}$. The force is given by $F=-dV/dx=2ax-4bx^3$. The angular frequency of small oscilations about the minimum can be found from, $\begin{eqnarray} F(x-x_0)&=&2a(x-x_0)-4b(x-x_0)^3\\ &\approx& 2ax-4b(3x_0^2x) + O(2)\\ &=&(2a-12bx_0^2)x\\ m\ddot{x}&=&-4ax\\ \ddot{x}&=&-\omega^2 x \Rightarrow \omega^2 = \frac{4a}{m} \end{eqnarray}$ where one might recall the binomial theorem or pascal's triangle to quickly figure out the trinomial coefficients. One finds that $\omega =2\sqrt{\frac{a}{m}}$, as in choice (D). Click here to jump to the problem!
 40 Click here to jump to the problem! GR8677 #93 Mechanics$\Rightarrow$}Potential The problem gives a nifty potential energy graph. The period is due to each part of the potential graph. For the simple harmonic oscillator (SHO) part, one remembers the formula $T=2\pi\sqrt{m/k}$ (to wit: $m\ddot{x}=-kx \Rightarrow \ddot{x}=-\omega^2 x \Rightarrow \omega =\sqrt{k/m}=2\pi f=2\pi/T \Rightarrow T=2\pi\sqrt{m/k}$). However, that period is for a particle to oscillate from one end of the potential curve to the other end and then back again. Since the graph shows only half of the usual SHO potential, the period contribution from the SHO part should be half the usual period: $T_{SHO_{1/2}}=\pi\sqrt{m/k}$ For the gravitational potential, one can calculate the period from the usual kinematics equation for constant acceleration. Recall the baby-physics equation, $x=\sqrt{1}{2}g t^2 \Rightarrow t=\sqrt{2x/g}$. The quantity needs to be converted to the relevant parameters of the problem. The problem supplies the constraint that the energy is constant, $E=\frac{1}{2}mv^2+mgx$. At the endpoint, one has $v=0\Rightarrow x=E/mg$. Plugging this into the equation for time, one gets $t=\sqrt{2E/mg^2}$. Since the particle has to travel from the origin to the right endpoint and then back to the origin, the total time contribution from this potential is twice that, $T_{grav}=2\sqrt{2E/mg^2}$. The total period is thus the sum of the above contributions, which is choice (D). Click here to jump to the problem!
 41 Click here to jump to the problem! GR8677 #1 Mechanics$\Rightarrow$}Acceleration Vector This is a conceptual vectors problem. Tangential: One knows that the velocity of the bob at the endpoints is 0, and that it has maximal acceleration (pulling it back towards the center) at the endpoints. In the center, it has maximal velocity and thus minimal acceleration. Normal: The normal acceleration varies from 0 at the endpoints to its maximal value at the center. The normal acceleration varies so that the sum of tangential and normal accelerations is a constant. Click here to jump to the problem!
 42 Click here to jump to the problem! GR8677 #2 Mechanics$\Rightarrow$}Centripetal Force Frictional force is given by $F_f = \mu N$, where N is the normal force. Centripetal force gives the net force to be, $\sum F = mv^2/r = \mu mg$. The m's cancel out, and one gets $r=v^2/(\mu g)$. The revolutions per minute $\omega$ given in the problem can be converted to velocity by $v=r\omega=r(33.3 rev/min)(2Pi rad/rev)(min/60s)\approx \pi r m/s$. Plugging this into the equation for r above, one has $r= \pi^2r^2/(\mu g)\Rightarrow r=\mu g/pi^2 \approx 3/\pi^2 <\approx 1/3$, which would be choice (D). Click here to jump to the problem!
 43 Click here to jump to the problem! GR8677 #3 Mechanics$\Rightarrow$}Kepler's Third Law Recall Kepler's Third Law stated in its most popular form, The square of the period is proportional to the cube of the orbital radius. (Technically, the orbital radius is the semimajor axis of the ellipse.) Recast that commonsense fact above into equations to get $T \propto R^{3/2}$, as in choice (D). Click here to jump to the problem!
 44 Click here to jump to the problem! GR8677 #4 Mechanics$\Rightarrow$}Conservation of Momentum From conservation of momentum, one has $2mv = 3mv' \Rightarrow v'=2/3 v$. The initial kinetic energy is $1/2 m v^2 = mv^2$. The final kinetic energy is $1/2 (3m) v'^2 = 3/2 m (2/3 v)^2$. The ratio of the final to initial kinetic energy is $2/3$. The kinetic energy lost in the collision is 1 minus that ratio. Thus, the answer is $1/3$, as in choice (C). Click here to jump to the problem!
 45 Click here to jump to the problem! GR8677 #14 Mechanics$\Rightarrow$}Cross Section Formula When the detector is placed near the source, the particles enter it directly from the circle at the end of the cylinder. The area is $A=\pi d^2/4$ When the detector is placed 1 m away, the particles enter it through a sphere of radius 1 m, thus $A=4\pi R^2 = 4\pi$. The ratio of areas gives the fraction of detected gamma rays, $d^2/16=8^2E-4/16=4E-4$, as in choice (C). This is due to Jon Jockers. Click here to jump to the problem!
 46 Click here to jump to the problem! GR8677 #22 Mechanics$\Rightarrow$}Orbits The planet orbits in an ellipse, and thus the sum of the minimum and maximum distance is twice the semimajor axis of the orbit. Choice (E) is out. From Kepler's third law (which can be derived from the formalism of central force theory), one has $T^2\propto r^3$. Thus, knowing the distances, one can find the period. Choice (D) is out. The speed can be found from conservation of energy. For an elliptical orbit, one has $E=-k/(2a) = 1/2 mv^2 - k/r$, where $k=GmM$ and a is the semimajor axis. (One can find the total energy of the orbit from the Virial Theorem. Recall that $\langle T \rangle = -\langle \sum F\cdot r\rangle/2$. Plug in the gravitational force $F=k/r^2$ to get $\langle T \rangle = \langle k/2r \rangle = \langle k/2a \rangle$. Thus, $T+V = k/2a - k/a = -k/2a$.) The full form of Kepler's third law is $\tau^2 \approx a^3/(GM)$. Thus, one can determine the mass of the planet. The only one that remains is choice (A). Click here to jump to the problem!
 47 Click here to jump to the problem! GR8677 #23 Mechanics$\Rightarrow$}Vectors If a particle is moving in a circle, then its acceleration points towards the center---even if it is moving at a constant (magnitude) velocity. This is the normal component of acceleration. Now, if its tangential velocity is increasing, then it has a tangential acceleration. The tangential acceleration is given as just $a_t=10 m/s^2$. The normal acceleration is given by the centripetal acceleration formula $a_n=v^2/r = 100^2/10=10$. Vectorially add the normal and tangential acceleration and dot it with the tangential velocity to find that the angle between them is just 45 degrees. Click here to jump to the problem!
 48 Click here to jump to the problem! GR8677 #24 Mechanics$\Rightarrow$}Kinematics If a stone is thrown at such an angle at an initial velocity, its horizontal $v_x$ vs t graph should be constant and positive $v_x=v_{x0}=v_0\cos(45^\circ)$. Thus, choices (A) and (E) are out. Recalling the basic kinematics equation $v_y=v_{y0}-gt$, one eliminates choice (D), since that shows a parabolic time dependence, when a linear one is required. Since the slope is negative, the $v_y$-graph should look like III one has choice (C). (If one forgets the basic equations above, one can derive it all from summing up the net force $\ddot{y} = -g$. Integrate both sides to get velocity. Integrate again to get position.) Click here to jump to the problem!
 49 Click here to jump to the problem! GR8677 #25 Mechanics$\Rightarrow$}Moment of Inertia The moment of inertia of the center penny about the center is just $1/2 m r^2$ The moment of inertia of any one of the other pennis about the center is given by the parallel axis theorem, $I=I_{CM} + md^2$, where d is the distance from the new point from the center of mass. $I_{CM}=mr^2/2$ for each penny, and thus one has $I=mr^2/2+md^2=mr^2/2+m(2r)^2=9/2mr^2$, since the distance from the center of each penny to the center of the configuration is 2r. Since there are 6 pennies on the outside, one has the total inertia $I=1/2mr^2+54/2mr^2=55/2mr^2$, as in choice (E). Click here to jump to the problem!
 50 Click here to jump to the problem! GR8677 #26 Mechanics$\Rightarrow$}Conservation of Energy Conservation of energy gives, $E=MgL/2=1/2 Mv^2 +1/2I\omega^2$, where the potential energy is given about the center of mass. The moment of inertia is about the center of mass, too. The translational velocity is the velocity at the center of mass. For a rod, one has $I=1/12 M L^2$. $v=\omega L/2$ Plug stuff in to get $MgL/2=1/2 M\omega^2/4 +1/2(1/12 M L^2)\omega^2=1/2(1/4+1/12)ML^2\omega^2 \Rightarrow g = (1/4+1/12)L\omega^2=(1/3)\omega^2 \Rightarrow \omega=\sqrt{3g/L}$. Now that one has the angular velocity, one can calculate the velocity at the endpoint; the rod rotates about its other end on the ground. $v=\omega \times r = L\omega = \sqrt{3gL}$, which is choice (C). Click here to jump to the problem!
 51 Click here to jump to the problem! GR8677 #54 Mechanics$\Rightarrow$}Impulse Impulse is defined as $J=\int F dt$. But, for this problem, one doesn't have to evaluate a messy integral. Instead, the area under the curve is just the sum of two triangles. $J=2\frac{1}{2}(2)(1)=2kgm/s$, as in choice (C). Click here to jump to the problem!
 52 Click here to jump to the problem! GR8677 #55 Mechanics$\Rightarrow$}Momentum From conservation of momentum in the horizontal direction, one has $mv = 2 m v' \cos\theta$. Solving for the final velocity, one has $v'=\frac{v}{2\cos\theta}$. Since $|\cos\theta| < 1$ for $\theta > 0$, one finds that $v' > v/2$, as in choice (E). Click here to jump to the problem!
 53 Click here to jump to the problem! GR8677 #56 Mechanics$\Rightarrow$}Fluid Statics The physical equation required is the buoyancy equation or Archimedes' Principal. The buoyant force is given by, $F_{buoyant} = \rho_{fluid} g V$. $\rho_{fluid}$ refers to the density of the fluid that buoy's the object. In this case, $\rho_{fluid}=\rho_a$ is just the density of air (not Helium). The buoyant force for this problem is $F_b = \rho_a g V_{He}$. This force must balance the load carried by the balloon. One now has (approximately, to simplify the calculations, since one is neglecting the weight of Helium) a simple sum of the forces problem, $\sum F = F_b - mg = 0 \Rightarrow 300g = \rho_a g V_{He}$. Now, solving for the volume of Helium required, one has $V_{He} = 300/\rho_a$. Since $\rho_a = 1.29$, this is about $230 m^3$. However, since the balloon has to lift the weight of the Helium as well, the actual volume should be slightly larger. The closest choice is (D). Click here to jump to the problem!
 54 Click here to jump to the problem! GR8677 #57 Mechanics$\Rightarrow$}Dimensional Analysis The only choice that has the right unit for force is choice (A), since the expression has the units $(kg/m^3) (m/s)^2 (m^2)=N$. Click here to jump to the problem!
 55 Click here to jump to the problem! GR8677 #72 Mechanics$\Rightarrow$}Forces Before the string is cut, one has the basic static equilibrium condition, for each block $\sum F = m_u a = 0 = T - m_u g - kx$ and $\sum F = m_b a = 0 = kx - m_b g$, where $m_u$ refers to the upper block and $m_b$ refers to the bottom block. Adding, one has $(m_u + m_b) a = 0 = T-m_ug - m_b$. After the string is cut, the tension goes to 0, but one has a non-zero net acceleration. The top mass has $\sum F = -m_u a = -m_ug - kx$, where the minus sign comes in since the question wants the downward acceleration. The bottom mass has $\sum F = m_b a =kx-m_b g$. But, immediately after the string is cut, the lower mass has 0 acceleration. Thus, $kx=m_b g$. Plugging this into the equation for the upper mass, one finds that $a = 2g$, as in choice (E). Click here to jump to the problem!
 56 Click here to jump to the problem! GR8677 #73 Mechanics$\Rightarrow$}Forces The net force of the system is $\sum F = (m_a+m_b) a$, and thus the net acceleration due to this force is $a = F/(m_a+m_b)$. The net force for mass A is just $m_a a = F - F_{ab}$. By Newton's Third Law, $F_{ab}=N$ is just the normal force exerted by A on B. Solving for the normal force, one finds that $N = F - m_a a = F(1 - m_a/(m_a+m_b))=F(m_b/(m_a+m_b))$. Summing up the vertical forces on mass B, and using the fact that the frictional force is just $f=\mu N$, one finds that $\sum F_y = \mu N-m_b g=0$ for the applied force to balance its mass completely. Thus, $F= (m_a+m_b)g/\mu$, which is approximately 400N, as in choice (D) after plugging in the numbers. Click here to jump to the problem!
 57 Click here to jump to the problem! GR8677 #74 Mechanics$\Rightarrow$}Lagrangians The Lagrangian equation of motion is given by $\frac{\partial L}{\partial q} = \frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$ for the generalized coordinate q. Chunking out the derivatives, one finds that $\frac{\partial L}{\partial q}=4bq^3$ $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=2a\ddot{q}$ Setting the two equal to each other as in the Lagrangian equations of motion given above (without undetermined multipliers), one finds that $2bq^3=a\ddot{q}$, which gives choice (D). Click here to jump to the problem!
 58 Click here to jump to the problem! GR8677 #75 Mechanics$\Rightarrow$}Transformations Recall the matrix equation for a rotational transformation of a coordinate system, $R(\theta)$= $\begin{eqnarray} \cos\theta & & \sin\theta \\ -\sin\theta & & \cos\theta \end{eqnarray}$ Equating coefficients, one has $\sin\theta = \sqrt{3}/2$. Thus, $\theta = \pi/3$, which corresponds to a 60-degree counter-clockwise rotation, as in choice (E). (Recall the unit circle.) Click here to jump to the problem!
 59 Click here to jump to the problem! GR8677 #89 Mechanics$\Rightarrow$}Conservation of Angular Momentum The initial angular momentum is $L_0 = (MR^2/2+mR^2) \omega = (M/2+m) R^2\omega$ The final angular momentum is $L_f = (MR^2/2) \omega^{'}$, since the radius of gyration of the young padawan is now 0. Conservation of angular momentum requires that $L_0=L_f \Rightarrow \omega^{'} = (M/2+m)/(M/2) = 2.8rad/s$. Click here to jump to the problem!
 60 Click here to jump to the problem! GR8677 #90 Mechanics$\Rightarrow$}Normal Modes Note that $\omega =\sqrt{\frac{k_{eff}}{m_{eff}}}$: For figure 1, the potential energy is $U=2\times 1/2kx^2=kx^2$. The kinetic energy is just $T=1/2 m\dot{x}^2$. Thus, $k_{eff} = 2k$ and $m_{eff}=m$. Thus, $\omega_1 = \sqrt{2k/m}$. For figure 2, the potential energy is $U=2\times 1/2 k (x/2)^2$, since each spring travels only half as far. The kinetic energy is the same as in figure 1. Thus, $k_{eff}=k/2$ and $m_{eff}=m$. So, one has, $\omega_2 =\sqrt{k/2m}$ Since $\omega = 2\pi f = 2\pi/T$, the period $T_1/T_2 = \omega_2/\omega_1 = \sqrt{(1/2)/(2)}=1/2$, as in choice (A). Click here to jump to the problem!
 61 Click here to jump to the problem! GR8677 #91 Mechanics$\Rightarrow$}Conservation of Energy Conservation of energy gives $Mg(H) = \frac{1}{2}Mv^2 + \frac{1}{2}I \omega^2$. $v=R\omega$, and thus the equation becomes $Mg(H) = \frac{1}{2}Mv^2 + \frac{1}{2}I (v/R)^2$. Thus, $I=2(MgH - Mv^2)R^2/v^2$. (Note that the height of the center of mass is the same at both the end and the start, thus the extra bit of the potential energy MgR cancels out. Thanks to the user keflavich for this correction.) Given $v^2 = 8gH/7$, one plugs it into the equation above to get $I=2(MgH -M4gH/7)R^2/(8gH/7)=2(3MgH/7)R^2/(8gH/7)=3MR^2/4$, as in choice (B). Click here to jump to the problem!
 62 Click here to jump to the problem! GR8677 #92 Mechanics$\Rightarrow$}Hamiltonian The Hamiltonian is just the sum of the kinetic and potential energy, $H=T+V$. The kinetic energy due to each mass is $T_i = p_i^2/(2m)$. The potential energy is just $U=1/2 k (\Delta l)^2$. $\Delta l = l -l_0$, and thus factoring out the $1/2$, one arrives at choice (E). Click here to jump to the problem!
 63 Click here to jump to the problem! GR8677 #4 Mechanics$\Rightarrow$}Gravitational Law Recall the famous inverse square law determined almost half a millennium ago, $F=\frac{k}{r^2}, $ where $k=GMm$. The ratio of two inverse-square forces ($r>R$, where $R$ is the radius of the planet or huge heavy object) would be $\frac{F(r_1)}{F(r_2)}=\frac{4r_2^2}{r_1^2}.$ Thus, $\frac{F(R)}{F(2R)}=\frac{4R^2}{R^2}=4$, which is choice (C). Click here to jump to the problem!
 64 Click here to jump to the problem! GR8677 #5 Mechanics$\Rightarrow$}Gauss Law The inverse-square law doesn't hold inside the Earth, just like how Coulomb's law doesn't hold inside a solid sphere of uniform charge density. In electrostatics, one can use Gauss Law to determine the electric field inside a uniformly charged sphere. The gravitational version of Gauss Law works similarly in this mechanics question since $\nabla \cdot \vec{E}=\rho_e\Rightarrow\nabla\cdot\vec{g}=\rho_M$, where $\rho_M$ is the mass density of $M$. In short, the gravitational field $\vec{g}$ plays the analogous role here as that of $\vec{E}$ Thus, $\oint \vec{g}\cdot d\vec{a}=\int \rho dV$. So, for $r, $g(4\pi r^2)=\rho \frac{4}{3}\pi r^3 \Rightarrow g= r\frac{\rho}{3}$, where one assumes $\rho$ is constant. To express the usual inverse-square law in terms of $\rho$, one can apply the gravitational Gauss Law again for $r>R$, $g(4\pi r^2)=\rho \frac{4}{3}\pi R^3 \Rightarrow g=\frac{R^3}{r^2}\frac{\rho}{3}$ Since $\vec{F}=m\vec{g}$ Therefore, $\frac{F(R)}{F(R/2)}=2.$ which is choice (C). Click here to jump to the problem!
 65 Click here to jump to the problem! GR8677 #6 Mechanics$\Rightarrow$}Method of Sections By symmetry, one can analyze this problem by considering only one triangular wedge. The normal force on one wedge is just $N=(m+M/2)g$, since by symmetry, the wedge (m) carries half the weight of the cube (M). The frictional force is given by $f=\mu N = \mu (m+M/2)g$. Sum of the forces in the horizontal-direction yields $F_x=0\leq f - N_M/\sqrt{2} = \mu(m+M/2)g - Mg/2$ for static equilibrium to remain valid. (Note that the normal force of the cube is given by $N_M=Mg/\sqrt{2}$ since, summing up the forces perpendicular to the plane for M, one has, $N_M\sin(\pi/4)=Mg/2$. Also, note that it acts at a 45 degree angle to the wedge.) Solving, one has $\mu(m+M/2)g \geq Mg/2 \Rightarrow M \leq \frac{2\mu m}{1-\mu}$. (In a typical mechanical engineering course, this elegant method by symmetry is called the method of sections.) Click here to jump to the problem!
 66 Click here to jump to the problem! GR8677 #7 Mechanics$\Rightarrow$}Normal Modes For normal mode oscillations, there is always a symmetric mode where the masses move together as if just one mass. There are three degrees of freedom in this system, and ETS is nice enough to supply the test-taker with two of them. Since the symmetric mode frequency is not listed, choose choice it!---as in (A). Click here to jump to the problem!
 67 Click here to jump to the problem! GR8677 #8 Mechanics$\Rightarrow$}Torque The problem wants a negative $z$ component for $\tau$. Recall that $r \times F=0$ whenever $r$ and $F$ are parallel (or antiparallel). Thus, choices (A), (B), (E) are immediately eliminated. One can work out the cross-product to find that (D) yields a positive $\tau_z$, thus (C) must be it. Click here to jump to the problem!
 68 Click here to jump to the problem! GR8677 #19 Mechanics$\Rightarrow$}Mass of Earth If one does not remember the mass of the earth to be on the order of $10^{24} kg$, one might remember the mass of the sun to be $10^{30} kg$. Since the earth weighs much less than that, the answer would have to be either (A) or (B). The problem gives the radius of the earth, and one can assume that the density of the earth is a few thousand $kg/m^3$ and deduce an approximate mass from $m=\rho V$. The answer comes out to about $10^{22}$, which implies that the earth is probably a bit more dense than one's original assumption. In either case, the earth can't be, on average, uniformly $10^9 kg/m^3$ dense. Thus (A) is the best (and correct) answer. An alternate solution is provided by the user SlickAce21. Equating the mass of some object with the gravitational force, one has $mg=GmM/r^2\Rightarrow g = GM/r^2 \Rightarrow M = gr^2/G \approx 6 E24$. Click here to jump to the problem!
 69 Click here to jump to the problem! GR8677 #40 Mechanics$\Rightarrow$}Centripetal Force There is no tangential acceleration (since otherwise it would slide and not roll---the frictional force balances the forward acceleration force). However, there is a centripetal acceleration that pulls the particles back in a circle, as in choice (C). This acceleration propels the tangential velocity to continue spinning in a circle. Click here to jump to the problem!
 70 Click here to jump to the problem! GR8677 #41 Mechanics$\Rightarrow$}Energy The kinetic energy is related to the inertia I and angular velocity $\omega$ by $K = \frac{1}{2}I\omega^2$. The problem supplies $I=4kgm^2$ so one needs not calculate the moment of inertia. The angular velocity starts at $80 rad/s$ and ends at $40 rad/s$. Thus, the kinetic energy lost $\Delta K = \frac{1}{2}I(\omega_f^2-\omega_0^2)=\frac{1}{2}(4)(40^2-80^2)=2(1600-6400)=-9600J$, as in choice (D). Click here to jump to the problem!
 71 Click here to jump to the problem! GR8677 #42 Mechanics$\Rightarrow$}Angular Kinematics Kinematics with angular quantities is exactly like linear kinematics with $x \rightarrow \theta$ (length to angle) $a \rightarrow \alpha$ (linear acceleration to angular acceleration) $v \rightarrow \omega$ (linear velocity to angular velocity) $m \rightarrow I$ (mass to moment of inertia) $F \rightarrow \tau$ (force to torque). Thus, one transforms $v=v_0+at \Rightarrow \omega=\omega_0+\alpha t$. Plugging in the given quantities, one gets $\alpha = \frac{\omega-\omega_0}{t}=\frac{40-80}{10}=-4rad/s^2$. The torque is given by $\tau = I \alpha = -16Nm$, whose magnitude is given by choice (D). Click here to jump to the problem!
 72 Click here to jump to the problem! GR8677 #43 Mechanics$\Rightarrow$}Lagrangians Recall the Lagrangian equations of motion $\frac{\partial L}{\partial q} =\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$. If $\frac{\partial L}{\partial q}=0$ then $\frac{\partial L}{\partial \dot{q}}=constant$, since its time-derivative is 0. One can relate energy to momentum from elementary considerations by $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$, where L is the kinetic energy $0.5m\dot{x}^2$. Thus, the generalized momentum defined for a generalized coordinate is just $p_n = \frac{\partial L}{\partial \dot{x}}$. From the above deductions, the generalized momentum is constant, as in choice (B). (Incidentally, the ignorable or cyclic coordinate would be $q_n$ and not $p_n$ since it does not appear in the Lagrangian.) Click here to jump to the problem!
 73 Click here to jump to the problem! GR8677 #44 Mechanics$\Rightarrow$}Lagrangians The kinetic energy, in general, is given by $T=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)$. The potential energy is just $V=mgy$. The Lagrangian is given by $L=T-V=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)-mgy$. Now, given the constraint $y=ax^2$, one can differentiate it and plug it into the Lagrangian above to reexpress the Lagrangian in terms of just y, for example. Differentiating, one has $\frac{d}{dt}(y=ax^2) \Rightarrow \dot{y}=a2x\dot{x} \Rightarrow \dot{x}=\dot{y}/(2ax)$. Square that to get $\dot{x}^2=\frac{\dot{y}^2}{4a^2x^2}=\frac{\dot{y}^2}{4ay}$, where one replaces the $x^2$ through the given relation $y=ax^2$. Plug that back into the Lagrangian above to get exactly choice (A). Click here to jump to the problem!
 74 Click here to jump to the problem! GR8677 #45 Mechanics$\Rightarrow$}Conservation of Energy Conservation of energy gives $mgh = \frac{1}{2}mv_0^2$, where $v_0$ is the velocity of the ball before it strikes the ground. Thus, $v_0^2 = 2gh$. Afterward, the ball bounces back up with $v^{'}=0.8v$. Apply conservation of energy again to get $mgh^{'} = \frac{1}{2}mv^{'2} \Rightarrow v^{'2} = 2gh^{'} \Rightarrow h^{'}=v^{'2}/(2g)$. Plugging in $v^{'2}=0.8^2v_0^{2}$, one has $h^{'}=0.64h$, which is choice (D). Click here to jump to the problem!
 75 Click here to jump to the problem! GR8677 #61 Mechanics$\Rightarrow$}Small Oscillations One can derive the frequency of small oscillation for a rigid body in general by using the torque form of Newton's Laws: $\tau = I \ddot{\theta} = \vec{r} \times \vec{F}$. (I is moment of inertia, r is moment arm) In this case, one has a constant downwards force $F=mg$, which acts at a moment arm angle $\theta$. Thus, $I \ddot{\theta} = -rmg \sin\theta \approx -rmg \theta$, where the approximation works if $\theta << 1$. The equation of motion for small angles is thus $\ddot{\theta}=-(mgr/I)\theta$. This is similar in form to that of a simple harmonic oscillator with the angular frequency being $\omega=\sqrt{mgr/I}$. Now, the problem is to find the angular frequency for each system. One needs not worry about the rod, since it is massless and has no moment of inertia. The moment of inertia of system I is just $I_I = 2mr^2$. The radius of gyration r is just $r_I = 2r$ (an r for each mass). The moment of inertia of system II is $I_{II}=mr^2 + m (r/2)^2 = \frac{5}{4}mr^2$. The radius of gyration r is just $r_{II} = r/2+r=3r/2$. Thus $\omega_{II}/\omega_{I} = \sqrt{(r_{II}I_{I})/(r_{I}I_{II})}=\sqrt{2\times 3 r/2/(2(5/4)\times r)}=\sqrt{6/5}$, as in choice (A). Click here to jump to the problem!
 76 Click here to jump to the problem! GR8677 #66 Mechanics$\Rightarrow$}Work Work is defined by $W=\vec{F} \cdot d\vec{l}$. The force here is just due to gravity, thus $F=\rho y g$, where $\rho=2kg/m$ is the density of the chain. The chain is wound upwards, so work is $W=\int_0^{10} \rho g y dy = \frac{2}{2}(gx^2)^{10}_0=10\times 100 =1000J$, as in choice (C). (The approximation $g\approx 10m/s^2$ is made.) Click here to jump to the problem!
 77 Click here to jump to the problem! GR8677 #74 Mechanics$\Rightarrow$}Small Oscillations The small oscillations of the hoop has the same frequency as that of a simple pendulum. Thus, $\omega^2 = \frac{g}{l}$. However, in this case, $l$ is the distance from the center of mass to the oscillation point---which is just the radius of the loop. Since $\omega = 2\pi/T$, the period $T\propto \sqrt{\frac{r}{g}} \propto \sqrt{r}$ does not depend on mass. Since $r_x = 4 r_y$. The ratio of periods is $T_X/T_Y ~ T/\sqrt{r_y} ~ T/\sqrt{r_x/4} ~ 2$. Thus, the period of Y is just $T/2$. (Note, the technique of leaving out constants requires that $~$'s are used instead of $=$'s. Practice a few times with this technique, as this will save time on the actual exam.) Click here to jump to the problem!
 78 Click here to jump to the problem! GR8677 #78 Mechanics$\Rightarrow$}Multiple Particles The angular momentum equation gives the angular frequency. $L =I \omega = m\vec{r} \times \vec{v}$, which relates the angular momentum to the moment of inertia, the angular velocity, the radius of gyration and the linear velocity. The system spins about its center of mass, which is conserved. Since the pole is massless and the skaters are off the same mass, $r_{cm}=b/2$. The moment of inertia of the system is just $I=2mr_{cm}^2=2m(b/2)^2$. Thus, the angular momentum equation gives $I\omega = m\vec{r} \times \vec{v} \Rightarrow 2m(b/2)^2 \omega =m b/2 (v + 2v) \Rightarrow 2(b/2) \omega = 3v \Rightarrow \omega = 3v/b$, since the cross-products point in the same direction. Now that one has the angular velocity, one eliminates all but choices (B) and (C). Now, take the time-derivative of x for choices (B) and (C), then evaluate it at $t=0$. For B, one has $dx/dt = v + 1.5v \cos(3vt/b) \rightarrow 2.5v$ for $t=0$. For C, one has $dx/dt = 0.5v + 1.5v \cos(3vt/b) \rightarrow 2v$ for $t=0$. Since the top skater is initially at $v(0)=2v$, only choice (C) has the right initial condition. Choose choice (C). (FYI: The center of mass velocity is given by $v_{cm}=\frac{2mv-mv}{2m}=v/2$. One can also arrive at (C) by noting conservation of center of mass velocity, since there is no net force.) Click here to jump to the problem!
 79 Click here to jump to the problem! GR8677 #82 Mechanics$\Rightarrow$}Torque The problem gives $H=\int \tau dt = I \alpha t$, but $\omega = \alpha t$. Thus, $\omega = H/I$. The moment of inertia of a plate about the z-axis is just $1/3 Md^2$. Plug this into $\omega$ to get choice (D). Click here to jump to the problem!
 80 Click here to jump to the problem! GR8677 #87 Mechanics$\Rightarrow$}Conservation of Energy Do not immediately try applying the Virial Theorem to this one. Instead, consider conservation of energy. Coming in from far away, the particle has $E=V=0$ the total energy equal to the potential energy equal to 0. Alternatively, one has, for a circular orbit, the equality between centripetal force and the attractive force, $mv^2/r=K/r^3\Rightarrow v^2 = K/(mr^2)$. Thus, the kinetic energy is just $T=p^2/(2m) = K/(2r^2)$. Since $V = - \int Fdr = -K/(2r^2)$, where the extra negative sign for the potential energy is due to an attractive potential. Thus, the total energy $E=T+V = 0$, which is choice (C). (Alternate solution is due to user crichigno.)Click here to jump to the problem!
 81 Click here to jump to the problem! GR8677 #93 Mechanics$\Rightarrow$}Boundary Condition Getting low on time, one should begin scoring points based more of testing strategy than sound rigorous physics. At the initial release point, the acceleration is due to gravity and the tension is 0 (no centripetal acceleration). The only choice that gives $a(\theta=0) = g$ is choice (E). Click here to jump to the problem!
 82 Click here to jump to the problem! GR8677 #100 Mechanics$\Rightarrow$}Sum of Moments Take the sum of the moments (or torque) about the triangular pivot fulcrum and set it to 0. $\sum M = 20gd +20gq - 40gx =0$, where $d$ is the distance from the fulcrum to the 20kg mass, $x$ is the distance from the fulcrum to the 40kg mass and $q$ is the distance from the fulcrum to the center of mass of the rod. From conservation of length, one also has $d+x=10$ and $q+x=5$. Plug everything into the moment equation. Shake and bake at 300 K. Solve for q to get choice (C). Alternatively, one can solve this problem in one fell swoop. Taking $q$ as the distance from the fulcrum to the center of mass of the rod, one sums the moment about the fulcrum to get $\sum M = 20gq + 20g(5+q) - 40(5-q)g =0$. Solve for $q$ to get choice (C). (This is due to the user astro_allison.)Click here to jump to the problem!

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