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GR9277 #42
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{42}
A cylinder with moment of inertia about a fixed axis initially rotates at 80 radians per second about this axis. A constant torque is applied to slow it down to 40 radians per second.

If the cylinder takes 10 seconds to reach 40 radians per second, the magnitude of the applied torque is

1. 80 Nm
2. 40 Nm
3. 32 Nm
4. 16 Nm
5. 8 Nm

Mechanics}Angular Kinematics

Kinematics with angular quantities is exactly like linear kinematics with

(length to angle)

(linear acceleration to angular acceleration)

(linear velocity to angular velocity)

(mass to moment of inertia)

(force to torque).

Thus, one transforms .

Plugging in the given quantities, one gets .

The torque is given by , whose magnitude is given by choice (D).  Alternate Solutions
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 mvgnzls2011-09-15 18:23:37 I did this solution: (is it correct)?:: L = IW T= = I = I dW/dt dW = 40 rad/s dt = 10 s I = 4 then T = 16 is this correct?Reply to this comment crossno2010-11-12 20:12:16 Just like linear impulse which says: Ft=P=mv angular impulse says t=I where I=4 =40 t=10 therefor =16Nm Reply to this comment jmason862009-09-03 18:52:53 This is similar to grae313's solution.. but here's what went through my head. F=ma T=I by analogy I = 4 (given) the change in (angular) velocity is 80-40=40 rad/sec Also given that the time it takes to do that change is 10 sec. By units, will give units of rad/s^2 which is an acceleration. So T = 4 * 40/10 = 4*4 = 16Reply to this comment grae3132007-10-14 14:16:38 Here's an alternate solution: The question asks for the magnitude. Reply to this comment      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$