A uniform 5.0 kg disk has a radius of 0.12 m and is pivoted so thatit rotates fr

Question

A uniform 5.0 kg disk has a radius of 0.12 m and is pivoted so thatit rotates freely about its axis. A string wrapped around the diskis pulled with a force equal to 20 N.

(a) What is the torque being exerted by this force about therotation axis?
(b) What is the angular acceleration of the disk?
(c) If the disk starts from rest, what is its angular speed after5.0 s?
(d) What is its kinetic energy after the 5.0 s?
(e) What is the angular displacement of the disk during the 5.0s?
(f) Show that the work done by the torque, ,equals the kinetic energy.

Explanation / Answer

(a)the torque being exerted by this force about the rotationaxis = F * r F = 20 N and r = 0.12 m (b)we know that = I * I is moment of inertia of the disk and is its angularacceleration or = (/I) = (/(m * r^2/2)) = (2/m *r^2) ------------(1) (c)the angular speed of the disk is w = wo + t wo= 0 or w = t the value of is obtained from equation (1) and t = 5.0s (d)the kinetic energy of the disk is K = (1/2)m * v^2 = (1/2)m * (r * w)^2 = (1/2)m * r^2 *w^2 (e)the angular displacement of the disk is = wot + (1/2)t^2 t = 5.0 s (f)In rotational motion,the work done by the torque is alwaysequal to the kinetic energy.This is because the torque is productof the force acting on the object(disk) and the perpendiculardistance from the center of the disk.The kinetic energy of theobject(disk) depends on the rotation of the object about an axispassing through the center of the disk.

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