Question

Tommy has a 20 kg metal disc with a redius 1.50 m. the disc started spinning from the rest by a constant horizontal force of 50.0 N applied tangentially to the metal disc. Find the kinetic energy of the metal disc after 4.3s.(Assume it is a solid cylinder.)

Answer 1

Mass o f the metal disc, M = 20 kg

Now, moment of inertia of the metal disc about the central axis, Mr = 1/2*M*(1.50m)^2

= 0.5*20*1.50^2 = 22.50 kg.m^2
Torque applied(T) = 50.0 N* 1.50m = 75 N.m

So, angular acceleration(a) = T/Mr = 75 N.m / 22.50 kg.m^2 = 3.33 rad/s^2
Final Angular Speed(w) after time t = 4.3s = a*t = 3.33 rad/s^2 *4.30 s = 14.32 rad/s

Therefore, the kinetic energy of the disc after 4.3 s -

E = 1/2*Mr*w^2 = 0.5*22.50*14.32^2 = 2306.6 Joule.