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  GR9277 #82
Problem
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\prob{82}
9277_82

A thin plate of mass M, length L, and width 2d is mounted vertically on a frictionless axle along the z-axis, as shown above. Initially the object is at rest. It is then tapped with a hammer to provide a torque $\tau$, which produces an angular impulse $\hat{H}$ about the z-axis of magnitude $H=\int \tau dt$. What is the angular speed $\omega$ of the plate about the z-axis after the tap?


  1. $\frac{H}{2Md^2}$
  2. $\frac{H}{Md^2}$
  3. $\frac{2H}{Md^2}$
  4. $\frac{3H}{2Md^2}$
  5. $\frac{4H}{Md^2}$
Mechanics}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).
See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
dberger8
2016-08-19 20:19:40		A lot like jmason86, I arrived at the answer a slightly different way. \\sum \\tau = I\\frac{d\\omega}{dt}=\\frac{dH}{dt} \\Rightarrow Id\\omega=dH \\Rightarrow \\omega=\\frac{H}{I}. Alternate Solution - Unverified
jmason86
2009-09-03 19:36:44		I arrived at \omega = \frac{H}{I} another way.
By analogy: Impulse=\int Fdt = \delta p = \delta (mv) and in the rotational system you get \delta (I\omega) instead. Since the initial angular velocity was 0, the delta just becomes I\omega = H. Some simple algebra and bam. 		Alternate Solution - Unverified
Comments
dberger8
2016-08-19 20:19:40		A lot like jmason86, I arrived at the answer a slightly different way. \\sum \\tau = I\\frac{d\\omega}{dt}=\\frac{dH}{dt} \\Rightarrow Id\\omega=dH \\Rightarrow \\omega=\\frac{H}{I}. Alternate Solution - Unverified
asa1985
2011-10-06 09:43:46		Torque=dL/dt;
where L is Angular Momentum;

Then H=\int_ Troque dt
H=\int_ dL
H=L
As we know L= Iw

w=L/I		NEC
kroner
2009-09-26 18:13:21		We know \omega will depend on I. The moment of inertia of a linear object like this is going to have a factor of 3 in it somewhere in terms of the mass and length, and there's no way for a factor of 3 to enter into the calculation elsewhere to cancel it out. That leaves choice (D).NEC
jmason86
2009-09-03 19:36:44		I arrived at \omega = \frac{H}{I} another way.
By analogy: Impulse=\int Fdt = \delta p = \delta (mv) and in the rotational system you get \delta (I\omega) instead. Since the initial angular velocity was 0, the delta just becomes I\omega = H. Some simple algebra and bam. 		Alternate Solution - Unverified
Anastomosis
2008-04-09 20:51:25		Although the test book doesn't give the moment of inertia for a plate, it does give it for a rod--if you look at it, a plate is just a thick rod, extended in the z-direction (the direction that has no bearing on the moment of inertia).

Anyway, I for a rod is \frac{1}{12}Ml^2, where l is just equal to 2d, or 2r in the general case.

So, \frac{1}{12}M(2d)^2 = \frac{1}{3}Md^2		NEC
FA
2007-04-13 00:29:06		typo alert. there should be no 2 in the denominator of the answer in D) NEC
i3taesun
2007-04-11 01:25:54		Like I=\int_0^\tFdt=delta P, H=\int_0^\ttorque dt=delta L.
Therefore, we should know the changing quantity of angular momentum.
NEC
cherianjudy
2006-11-03 16:11:02		NEC
jcain6
2005-11-23 07:00:05		I think w = at not a/t. This is why w = H/I right?
yosun
2005-11-23 15:02:18		 jcain6: thanks for the typo-alert; it has been corrected. (while typing up this solution, my pinky was evidently slash-key happy.)
Fixed Typos!

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A lot like jmason86, I arrived at the answer a slightly different way. \\sum \\tau = I\\frac{d\\omega}{dt}=\\frac{dH}{dt} \\Rightarrow Id\\omega=dH \\Rightarrow \\omega=\\frac{H}{I}.

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