A cylindrical rod 24.5 cm long with a massof 1.20 kg and a radius of 1.50 cm has

Question

A cylindrical rod 24.5 cm long with a massof 1.20 kg and a radius of 1.50 cm has a ball of diameter of7.80 cm and a mass of 2.00 kg attached toone end. The arrangement is originally vertical and stationary,with the ball at the top. The apparatus is free to pivot about thebottom end of the rod. (a) After it falls through 90°, what is itsrotational kinetic energy?
1 J

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

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

(d) How does this compare with the speed if the ball had fallenfreely through the same distance of 28.4cm?
vswing is 4---Select---greater thanless thanvfall by 5% (a) After it falls through 90°, what is itsrotational kinetic energy?
1 J

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

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

(d) How does this compare with the speed if the ball had fallenfreely through the same distance of 28.4cm?
vswing is 4---Select---greater thanless thanvfall by 5%

Explanation / Answer

Potential energy with respect to the axis of rotation: 1.2 * (.245 / 2 ) * g + 2 * (.245 + .078 / 2) g = .715 g= 7.01 J 1/2 I 2 = 7.01J     KE at 90 deg equals PE attop I = 1/3 * 1.2 * .2452 + 2 / 5 * 2 * (.078 /2)2 + 2 * (.245 + .078 / 2)2 = .187kg-m2 Using the parallel axis theorem to get inertia of ball aroundaxis of rotation Then = (2 * 7.01 / .187) = 8.67 / sec The linear speed of the ball v = (L + R / 2) =8.67 * (.245 + .078 / 2) = 2.46 m/s 1/2 m v2 = m g h    v = (2 gh) = (2 * 9.8 * .284) = 2.35 m/s Speed of swing > speed of ball by   (2.46 - 2.35)* 100 / 2.35 = 4.7 % Using the parallel axis theorem to get inertia of ball aroundaxis of rotation Then = (2 * 7.01 / .187) = 8.67 / sec The linear speed of the ball v = (L + R / 2) =8.67 * (.245 + .078 / 2) = 2.46 m/s 1/2 m v2 = m g h    v = (2 gh) = (2 * 9.8 * .284) = 2.35 m/s Speed of swing > speed of ball by   (2.46 - 2.35)* 100 / 2.35 = 4.7 %

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