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# Angular Velocity: Slender Bar, Stepped Disk and Slender Rods

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1. The slender bar with a mass of 4 kg is released from rest at an angle of 35 degrees. What is the angular velocity of the bar when it is vertical?

2. The stepped disk weighs 50 lb and its mass moment of inertia about its center is I = 0.4 slug-ft2. If it is released from rest, determine its angular velocity when it falls 3 ft. The rope does not stretch.

3. An assembly consists of two 15-lb slender rods and a 20-lb disk. The spring is unstretched when &#952; = 45Âº and the assembly is released from rest at this position. The disk rolls without slipping. Determine the angular velocity of rod AB at the instant &#952; = 0.

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#### Solution Preview

1.) mass of the rod m = 4 kg
length of the rod l = 2 m
the distance of the pivoted point from the center of the rod r = 0.5 m

Because initial angle of the rod with the vertical, theta = 35 deg
Finally, when the rod is vertical, angle = 0

Hence, loss in potential energy (U) = gain of rotational K.E.,
U = m*g*r*(1 - cos(theta)) = K = (1/2)*I*w^2
NOTE: potential energy is calculated due to change in position of the center of gravity of the rod = r - r*cos(theta)

moment of inertia of the rod about the pivoted point I = m*l^2/12 + m*r^2 = (4*2^2/12) + 4*0.5^2 = 2.33 kg.m^2
angular velocity w = ?
=> w = sqrt(2*m*g*r*(1 - cos(theta)/I)