GREPhysics.NET: GR9277 #42

GR9277 #42

Problem

GREPhysics.NET Official Solution

\prob{42}
A cylinder with moment of inertia $4kgm^2$ 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

80 Nm
40 Nm
32 Nm
16 Nm
8 Nm

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).

Alternate Solutions

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Comments

mvgnzls 2011-09-15 18:23:37	I did this solution: (is it correct)?:: L = IW T= $\dot{L}$ = I $\dot{W}$ = I dW/dt dW = 40 rad/s dt = 10 s I = 4 then T = 16 is this correct? Reply to this comment
crossno 2010-11-12 20:12:16	Just like linear impulse which says: F $\Delta$ t= $\Delta$ P=m $\Delta$ v angular impulse says $\tau$ $\Delta$ t=I $\Delta$ $\omega$ where I=4 $\Delta$ $\omega$ =40 $\Delta$ t=10 therefor $\tau$ =16Nm Reply to this comment
jmason86 2009-09-03 18:52:53	This is similar to grae313's solution.. but here's what went through my head. F=ma T=I $\alpha$ 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, $\frac{\delta V}{\delta t}$ will give units of rad/s^2 which is an acceleration. So T = 4 * 40/10 = 4*4 = 16 Reply to this comment
grae313 2007-10-14 14:16:38	Here's an alternate solution: $\ N =$ $\frac{dL}{dt} =$ $\frac{\Delta L}{\Delta t} =$ $\frac{I \Delta w}{\Delta t}$ $\ N =$ $\frac{4(40-80)}{10}$ $\ = -16$ The question asks for the magnitude. Reply to this comment

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Here's an alternate solution: $\ N =$ $\frac{dL}{dt} =$ $\frac{\Delta L}{\Delta t} =$ $\frac{I \Delta w}{\Delta t}$ $\ N =$ $\frac{4(40-80)}{10}$ $\ = -16$ The question asks for the magnitude.

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