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odie5533
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[Solved] Dynamics of Rotational Motion
A uniform disk has a mass of 3.4kg and a radius of 0.53m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 50rpm. A thin-walled hollow cylinder has the same mass and radius as the disk. It is released from rest, just above the turntable, and on the same vertical axis. The hollow cylinder slips on the turntable for 0.20s until it acquires the same final angular velocity as the turntable. The average torque exerted on the hollow cylinder during the 0.20s time interval in which slipping occurs is closest to:
A) 13 N m
B) 16 N m
C) 6.3 N m
D) 8.3 N m
E) 17 N m
Moment of inertia of a solid cylinder: [tex]\frac{1}{2}MR^2[/tex]
Moment of inertia of a thin-walled cylinder: [tex]MR^2[/tex]
[tex]50 rpm * \frac{2 \pi}{60} = 5.236 rad/s[/tex]
[tex]K_{1} = 0[/tex]
[tex]K_{2} = \frac{1}{2}I_{c}\omega^2[/tex]
[tex]K_{1} + W_{other} = K_{2}[/tex]
[tex]W = \frac{1}{2}I_{c}\omega^2[/tex]
[tex]W = \frac{1}{2}(MR^2)(5.236)^2[/tex]
[tex]W = \frac{1}{2}(3.4)(0.53^2)(5.236)^2 = 13.09 J[/tex]
I'm not sure how to relate this to average torque. I thought maybe W/time = torque, but that doesn't really work at all or make sense. W = t * d theta, but then I need theta for a slipping disk which seems hard to find.
[tex]\alpha = \frac{\Delta\omega}{\Delta t}[/tex]
[tex]\alpha = \frac{5.236}{0.20s} = 26.18 rad/s^2[/tex]
[tex]\tau = I\alpha[/tex]
[tex]\tau = (MR^2)(\alpha) = (3.4 * 0.53^2)(26.18)[/tex]
[tex]\tau = 25[/tex]
Which is nothing near any of the choices. I think I can't use the [tex]\tau = I\alpha[/tex].
Homework Statement
A uniform disk has a mass of 3.4kg and a radius of 0.53m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 50rpm. A thin-walled hollow cylinder has the same mass and radius as the disk. It is released from rest, just above the turntable, and on the same vertical axis. The hollow cylinder slips on the turntable for 0.20s until it acquires the same final angular velocity as the turntable. The average torque exerted on the hollow cylinder during the 0.20s time interval in which slipping occurs is closest to:
A) 13 N m
B) 16 N m
C) 6.3 N m
D) 8.3 N m
E) 17 N m
Homework Equations
Moment of inertia of a solid cylinder: [tex]\frac{1}{2}MR^2[/tex]
Moment of inertia of a thin-walled cylinder: [tex]MR^2[/tex]
The Attempt at a Solution
[tex]50 rpm * \frac{2 \pi}{60} = 5.236 rad/s[/tex]
[tex]K_{1} = 0[/tex]
[tex]K_{2} = \frac{1}{2}I_{c}\omega^2[/tex]
[tex]K_{1} + W_{other} = K_{2}[/tex]
[tex]W = \frac{1}{2}I_{c}\omega^2[/tex]
[tex]W = \frac{1}{2}(MR^2)(5.236)^2[/tex]
[tex]W = \frac{1}{2}(3.4)(0.53^2)(5.236)^2 = 13.09 J[/tex]
I'm not sure how to relate this to average torque. I thought maybe W/time = torque, but that doesn't really work at all or make sense. W = t * d theta, but then I need theta for a slipping disk which seems hard to find.
[tex]\alpha = \frac{\Delta\omega}{\Delta t}[/tex]
[tex]\alpha = \frac{5.236}{0.20s} = 26.18 rad/s^2[/tex]
[tex]\tau = I\alpha[/tex]
[tex]\tau = (MR^2)(\alpha) = (3.4 * 0.53^2)(26.18)[/tex]
[tex]\tau = 25[/tex]
Which is nothing near any of the choices. I think I can't use the [tex]\tau = I\alpha[/tex].
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