A thin, cylindrical rod = 22.6 cm long with a mass m = 1.20 kg has a ball of dia

Question

A thin, cylindrical rod = 22.6 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?

(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 27.6 cm?

Explanation / Answer

(a) PE is converted into KE. Measured relative to the pivot point, initial

PE = (1.20kg * 0.226m + 2.00kg * (0.0500 + 0.226)m) * 9.8m/s²

PE = 8.07 J becomes the KE at 90º

(b) KE = 8.07 J = ½I²

where

I = I_rod + I_sphere

I = mL²/3 + (2/5)MR² + M(R+L)² assuming the ball is solid

and invoking the parallel axis theorem

I = 1.20kg * (0.226m)² / 3 + (2/5) * 2.00kg * (0.0500m)² + 2.00kg * (0.276m)²

I = 0.175 kg·m²

so

8.07 J = ½ * 0.175kg·m² * ²

= 9.6 rad/s angular speed of rod and ball

(c) v = r = 9.6rad/s * (0.226 + 0.0500)m = 2.65 m/s

(d) V = (2gh) = (2 * 9.8m/s² * 0.276m) = 2.33 m/s

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