A uniform disk of mass M = 4.9 kg has a radius of 0.12 m and is pivoted so that it rotates freely about its axis. A string wrapped around the disk is pulled with a force F equal to 20 N.

(a) What is the torque being exerted by this force about the rotation axis?

(b) What is the angular acceleration of the disk?

(c) If the disk starts from rest, what is its angular speed after 4.6 s?

(d) What is its kinetic energy after the 4.6 s?

(e) What is the angular displacement of the disk during the 4.6 s?

(A) 2.4 N-m

(B)

(C) 315.426 rad/sec

(D) 1741.13 J

(E) 725.481 rad

Explanation:

We have given mass of the disk m = 4.9 kg

Radius r = 0.12 m, that is distance = 0.12 m

Force F = 20 N

(a) Torque is equal to product of force and distance

So torque , here F is force and r is distance

So

(B) Moment of inertia is equal to

So

Torque is equal to

So angular acceleration

(C) As the disk starts from rest

So initial angular speed

Time t = 4.6 sec

From first equation of motion we know that

So

(D) Kinetic energy is equal to

(E) From second equation of motion

the answer is a. at the equator