Question

A thin cylindrical rod =24.0 cm long with mass m=1.20kg has a ball of diameter d=8.00 cm and a mass M=2.00kg attatch to one end. The arrangement is originally verticaland stationary, with the ba;; at the top as shown in P10.50. The combination is free to pivot about the bottom end of the rod after being given a slight nudge. (a)After the combination roatates through 90 degress, what is its ratational kinetic energy?(b) What is the angular speed of the rod and ball? (c) What is the linearspeed of the center of the mass of the ball? (d) How does it compare with the speed had the ball fallen freely through the same distance of 28 cm?

Answer 1

L = 24.0 cm, m = 1.20 kg, radius of the ball r = 4.00 cm, M = 2.00 kg,
initial potential energy E = mg(L/2) + Mg(L + r)
After the combination rotates through 90 degrees,
its rotational kinetic energy = E = mg(L/2) + Mg(L + r) =6.899 J
the angular speed of the rod and ball is w
moment of inertia I = mL^2/3 + [2Mr^2/5 + M*(L + r)^2] (use parallel axis theorem)
I = 0.1926 kgm^2
E = Iw^2/2
so w = sqrt(2E/I) = 8.46 rad/s
linear speed of the v = (L+r)ω

=2.3688 m/s

d)The speed the ball would have if it had dropped a height L + r would be given by mg(L+r) = ( 1/2 mv^2

v = sqert 2g(L+r) = sqrt 2(9.8) ( 0.28 m/s) = 2.34 m/s