A thin, cylindrical rod = 23.8 cm long with a mass m = 1.20 kg has a ball of dia

Question

A thin, cylindrical rod = 23.8 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?
  J

(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 28.8 cm?

vswing is greater than less than  vfall by   

Explanation / Answer

a) initial potential energy of rod = M*g*l/2 = 1.2*9.81*0.119 = 1.400868 J

initial potential energy of bob = m*g*(l+d/2) = 2*9.81*(0.238+0.05) = 5.65056 J

final potential energy of bob and bar = 0J

change in potential energy =1.400868 J + 5.65056 J = 7.051428 J
The rotational kinetic energy = change in potential energy = 7.051428 J

b) RKE =[0.5*(omega)^2]* {(1.2*0.238^2)/3}+ {(2*2.0*(0.05)^2)/5} + (2.0*0.288^2)] = 7.051428 J , where omega is angular speed of the apparatus or
(omega)^2 = 74.013 or
omega = sq rt(74.013) = 8.603 rad/s

c)Linear speed, s of ball = 8.603*0.288 = 2.47768 m/s

d) then the speed, S would have been = sq rt(2*9.81*0.288) = 2.37709 m/s
s is slightly more than S exactly by = 0.1005 m/s

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