A 5.0 kg. 68-crn-dlameter cylinder rotates on an axle passing through one edge.

Question

A 5.0 kg. 68-crn-dlameter cylinder rotates on an axle passing through one edge. The axe is parallel to the floor. The cylinder is held with the center of mass at the same height as the axle, then released. (Figure 1) What is the magnitude of the cylinder's initial angular acceleration? Express your answer to two significant figures and include the appropriate units What Is the magnitude of the cylinder's angular velocity when it is directly below the axle? Express your answer to two significant figures and Include the appropriate units.

Explanation / Answer

ANSWER

Part A) Angular acceleration

Moment of Inertia = Icm = m r^2/2
Parallel Axis Theorem I = Icm + md^2
r = 34 cm
d = 34 cm
I = mr^2/2 + mr^2 = 3/2 mr^2
F = mg
Torque = = Fx d = mgr = I
= mgr/I = 2g/3r =2*9.8/3*0.34 = 19.22 rad/s2

Part B) Angular velocity

U = mgh = mgR = Krot = 1/2I^2 = 3mR^2*^2/4
= sqrt[4g/(3R)] = sqrt[4*9.8/(3*0.34)] = 6.199 rad/s

Regards!!!

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