GREPhysics.NET: GR9277 #82

GR9277 #82

Problem

GREPhysics.NET Official Solution

\prob{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?

$\frac{H}{2Md^2}$
$\frac{H}{Md^2}$
$\frac{2H}{Md^2}$
$\frac{3H}{2Md^2}$
$\frac{4H}{Md^2}$

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).

Alternate Solutions

dberger8 2016-08-19 20:19:40	A lot like jmason86, I arrived at the answer a slightly different way. $\\sum \\tau = I\\frac{d\\omega}{dt}=\\frac{dH}{dt} \\Rightarrow Id\\omega=dH \\Rightarrow \\omega=\\frac{H}{I}$ . Reply to this comment
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. Reply to this comment

Comments

dberger8
2016-08-19 20:19:40

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

Reply to this comment

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

Reply to this comment

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).

Reply to this comment

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.

Reply to this comment

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, $\frac{1}{12}M(2d)^2 = \frac{1}{3}Md^2$

Reply to this comment

FA
2007-04-13 00:29:06

typo alert. there should be no 2 in the denominator of the answer in D)

Reply to this comment

i3taesun
2007-04-11 01:25:54

Like I= $\int_0^\t$ Fdt=delta P, H= $\int_0^\t$ torque dt=delta L.
Therefore, we should know the changing quantity of angular momentum.

Reply to this comment

cherianjudy
2006-11-03 16:11:02

Reply to this comment

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.)

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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, $\frac{1}{12}M(2d)^2 = \frac{1}{3}Md^2$

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