33-A thin, cylindrical rod = 24.4 cm long with a mass m = 1.20 kg has a ball of

Question

33-A thin, cylindrical rod = 24.4 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. After the combination rotates through 90 degrees, what is its rotational kinetic energy? What is the angular speed of the rod and ball? What is the linear speed of the center of mass of the ball?

Explanation / Answer

Moment of Inertia of rod:

1/12 m (3 r^2 + l^2)

+ m(l/2)^2 ... parallel axis

I.rod = 1/12 m (3 r^2 + l^2) + m (l/2)^2 = 1/12 m (3 r^2 + 4 l^2)
[*Note: this will almost be the 1/3 m L^2 that you have, just a fraction more]
l = 24.4 cm, m = 1.2 kg

I.ball = 2/5 m r^2 + m (l + 1/2r)^2
[*Note: Again applying the parrallel axis theorem]
r = 3 cm, m = 2 kg

a. Set PE = KE
PE = mg(l/2) [rod] + mg (l + r) [ball]
= g (1.2 kg * (.244m/2) + 2kg (.274m))
= 6.812 J

b. PE = KE
6.812 J = 1/2 I w^2
w^2 = 2 * 6.812 J / ((1/12 m (3 r^2 + 4 l^2)[rod]) + (2/5 m r^2 + m (l + 1/2r)^2 [ball])
w^2 = 2 * 6.812 J / ((1/12 1.2kg * 4(.244m)^2) + 2/5 2kg (.03)^2 + 2kg (.274)^2)
w^2 = 2 * 6.812 J / .2459 kg m^2
w = 7.4424 rads/sec

c. The linear speed of the CENTER of the ball is

v/r = w
v = w r
v = w (.274m)
v = 2.039 m/s

d. Free fall
1/2 m v^2 = mgh
v = sqrt(2gh)
v = 2.318 m/s

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