A 5.0 kg, 60-cm-diameter disk rotates on an axle passingthrough one edge. The ax

Question

A 5.0 kg, 60-cm-diameter disk rotates on an axle passingthrough one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height asthe axle, then released. a) What is the cylinder's initial angular acceleration? b) What is the cylinder's angular velocity when it isdirectly below the axle? A 5.0 kg, 60-cm-diameter disk rotates on an axle passingthrough one edge. The axle is parallel to the floor. The cylinder is held with the center of mass at the same height asthe axle, then released. a) What is the cylinder's initial angular acceleration? b) What is the cylinder's angular velocity when it isdirectly below the axle?

Explanation / Answer

I = 1/2 M R2 + M R2 = 3 M R2/ 2       Moment of inertia byparallel axis theorem Torque = M g R = I = 3 M R2 /2 = 2 g / (3 R)    initial angularacceleration PE = M g R    initial potential energy KE = 1/2 I 2   rotational kineticenergy at bottom = 2 g (3 g / R) / 3   angularvelocity at bottom

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