Question

A thin, cylindrical rod ℓ = 22.2 cm long with a mass m = 1.20 kg has a ball of diameter d = 6.00 cm and mass M = 2.00 kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top as shown in the figure below. The combination is free to pivot about the bottom end of the rod after being given a slight nudge.

(a) After the combination rotates through 90 degrees, what is its rotational kinetic energy?

(b) What is the angular speed of the rod and ball?

(c) What is the linear speed of the center of mass of the ball?

(d) How does it compare with the speed had the ball fallen freely through the same distance of 25.2 cm?

Answer 1

Answer : convert all distances into meters and mass into kg if you need to.

L= 0.222 m , m= 1.20 kg, d= 0.06 m, M= 2.00 kg

Part (A)

1.Find the Potential Energy of the system. Use this equation for this problem only.
PE= mg(L/2) + Mg(L + d/2)
PE= (1.20)(9.80)(.222/2) + (2.00)(9.80)(.222+ .06/2)
PE= 1.30536 J + 4.9392 J
PE= 6.24456 J
That is also your rotational kinetic energy. (A)= 6.24456 J

For (B)

this will get hard to understand, but stick with it, and it will get you through this problem.
Need to find I_-total.
I_-rod= (1/3)ML^2= (1/3)(1.20)(.222^2)= 0.0197136
I_-ball= (2/5)MR^2+M(L+d/2)^2= (2/5)(2.0)(.03^2)+(2.0)[(.222+(.06/2))^2 =  0.127188
I_-total= 0.0197136+ 0.127188
I_-total= 0.1469016

Now that we have the I_-total of the system we can plug it up into the rotational kinetic energy.
K=(1/2)(I_-total)(w^2)
6.24456 J = (1/2) (0.1469016) (w^2)
6.24456 J = 0.0734508 (w^2)
Square root of (6.24456 /0.0734508) = w (angular speed of the ball and rod)
w= 9.22 rad/s is your answer to (B).

For (C)

you are worried about the ball:
v=w(L+d/2)
v= (9.22 rad/s)(.252 m)
v=2.32 m/s

For (D)

V = √(2gh) = √(2 x 9.8m/s² x 0.252m)

= 2.22 m/s