A thin, cylindrical rod ? = 26.0 cm long with a mass m = 1.20 kg has a ball of d

Question

A thin, cylindrical rod ? = 26.0 cm long with a mass m = 1.20 kg has a ball of diameter d = 10.00 cm and mass M = 2.00 kgattached 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?
J

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

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

(d) How does it compare with the speed had the ball fallen freely through the same distance of 31.0 cm?

vswing is (greater than or less than)   vfall by  %______.

Explanation / Answer

a) After the nudge, the system acquires a torque due to gravity force tau = r x mg whose magnitude is |tau| = rmg sin theta, where theta is the angle between mg and r.

The work made is W = |tau| Delta phi, where Delta phi is the angle by which it has rotated: in our case, it is 90 deg.

In our case, r = l + d/2 = 0.31 m, and m = 3.20 kg. Thus, W = W_fin - W_in =15.3 J - 0 = 15.3 J (at the beginning, there is no torque); for the kinetic energy theorem, this is equal to the kinetic energy acquired at the end of the rotation.

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