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

Question

A thin, cylindrical rod = 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? (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 25.6 cm? vswing is vfall by %.

Explanation / Answer

MOI of ball about its centre = 2/5*Mr^2

MOI of ball about pivot = 2/5*Mr^2 + M*(L+r)^2 = 2/5*2*((10*10^-2)/2)^2 + 2*[(20.6+10/2)*10^-2]^2 = 0.133072 kg-m^2

MOI of rod+ball about pivot I = mL^2/3 + 0.133072 = 1.2*(20.6*10^-2)^2/3 + 0.133072 = 0.15 kg-m^2

Position of centre of mass of assembly from pivot is given by y = [m(L/2) + M*(L+r)]/(m+M)

y = [1.2*(20.6/2)*10^-2 + 2*(20.6+10/2)*10^-2]/(1.2+2)

y = 0.198625 m

a)

Initial potential energy = (m+M)gy

Energy conservation: Final KE = (m+M)gy = (1.2+2)*9.81*0.198625 = 6.235 J

b)

Rotational KE = 1/2*I^2 = 6.235

1/2*0.15*^2 = 6.235

= 9.118 rad/s

c)

v = y = 0.198625*9.118 = 1.811 m/s

d)

For free fall, v' = [2g(L+r) = [2*9.81*(20.6+10/2)*10^-2] = 2.241 m/s

v/v' = 1.811/2.241 = 0.808 = 80.8%

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