A thin, cylindrical rod l = 20.6 cm long with a mass m = 1.20 kg has a ball of d

Question

A thin, cylindrical rod l = 20.6 cm long with a mass m = 1.20 kg has a ball of diameter d = 10.00 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? (in J)

(b) Calculate angular speed of the rod and ball (in rad/s).

(c) Calculate linear speed of the center of mass of the ball (in m/s).

(d) How does it compare with the speed had the ball fallen freely through the same distance of 25.6 cm?

Explanation / Answer

Moment of Inertia of rod:

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

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

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]
r = .5 cm, l = 20 .6cm, m = 1.2 kg

I.ball = 2/5 m r^2 + m (l + 1/2r)^2
[*Note: Again applying the parrallel axis theorem]
l = 20 .6cm, r = 5cm, m = 1.2kg

a. Set PE = KE
PE = mg(l/2) [rod] + mg (l + r) [ball]
= g (1.2 kg * (0.206/2) + 2kg (.256m))
= 6.22 J

b. PE = KE
6.22 J = 1/2 I w^2
w^2 = 2 * 6.22 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 * 6.22 J / ((1/12 1.2kg (3 (.05m)^2 + 4(.206m)^2) + 2/5 2kg (.05)^2 + 2kg (.256)^2)
w^2 = 2 * 6.22 J /0.15 kg m^2
w = 82.93rads/sec

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

v/r = w
v = w r


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

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