A spherical bowline ball encounters a 100 cm vertical rise on the way back to th

Question

A spherical bowline ball encounters a 100 cm vertical rise on the way back to the ball rack, as the figure 3 illustrates Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 3 m/s at the bottom of the rise. Find the translational speed at the top A tangential force of 15 N acts on the rim of a wheel 30 cm in diameter The wheel takes 2 s to complete its first revolution starting from rest. Determine the moment of inertia of the wheel.

Explanation / Answer

  Kinetic energy of the ball before the rise:
KE = m×v²/2

Change in potential energy of the ball after the rise:
PE = m×g×h

Change in kinetic energy after the rise:
KE = m×v²/2 - m×g×h = m×V²/2
V²/2 = v²/2 - g×h
V = (v² - 2×g×h)

V = [(3 m/s)² - 2×(9.8 m/s²)×(1 m)]
V = 3.25m/s

Torque = 15*0.3 = 4.5 N m
average angular velocity = (2*pi*0.3)/2 = 0.942 rad/s
angular velocity, 'omega' at the end of 2 s is given by (0+omega)/2 = 0.942 rad/s or
omega = 2*0.942 =1.884 rad/s angular acceleration, 'alpha' =1.884/2 = 0.942 rad/s^2
Moment of Inertia, MI = Torque /alpha = 4.5/ 0.942 = 4.77 kg m^2

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