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?
a. 0.667 rad/s
b. 1.06 rad/s
c. 0.888 rad/s
d. 1.74 rad/s
e. 2.31 rad/s
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.
a. 22.9 rad/s
b. 10.2 rad/s
c. 8.20 rad/s
d. 32.2 rad/s
e. 15.7 rad/s
3. An assembly consists of two 15-lb slender rods and a 20-lb disk. The spring is unstretched when θ = 45º 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 θ = 0.
a. 3.89 rad/s
b. 3.65 rad/s
c. 3.12 rad/s
d. 2.69 rad/s
e. 4.28 rad/s
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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)
=> w = sqrt(2*4*9.8*0.5*(1 - cos(35))/2.33) = 1.74 rad/s --Answer (d)
2.) mass of the disc, m = 50 lb
mass moment of inertia about its center, I = 0.4 slug.ft^2 = ...
Solution Summary
The answers are given step-by-step with some short explanation.