1. Find the total energy of the system at the given instant if the 20 kg disc has an angular velocity of 24 rad/s (clockwise), the bar AB has a mass of 2 kg, and the system is 1.5 meters above the ground. Assume the ground as the datum line (zero potential energy).

a. 353 J

b. 389 J

c. 294 J

d. 323 J

e. 429 J

2. A cylinder is pulled up a plane by the tension in a rope which passes over a frictionless pulley and is attached to a weight of 150 lb as shown. The cylinder weighs 100 lb and has a radius of 2 ft. The radius of the pulley is 4 ft. The cylinder moves from rest up a distance of 16 ft. What will be its speed?

1.) Find the total energy of the system at the given instant if the 20 kg disc has an angular velocity of 24 rad/s (clockwise), the bar AB has a mass of 2 kg, and the system is 1.5 meters above the ground. Assume

Total energy = Potential energy + K.E.
potential energy of the bar AB, Uab = m(bar)*g*h(bar) = 2*9.8*1.5 = 29.40 J
potential energy of the disc, Ud = m(disc)*g*h(disc) = 20*9.8*1.5 = 294.00 J
K.E. of the disc, K = 0.5*I(disc)*w^2

moment of inertia of the disc, I = 0.5*m(disc)*r^2 = 0.5*20*0.1^2 = 0.10 kg.m^2
angular velocity of the disc, w = 24 rad/s

Hence, K = 0.5*0.10*24 = 1.20 J

hence, total Energy = 29.40 + 294.00 + 1.20 = 324.6 == 323 J --Answer (d)

2.) For block: mass M = 150 lb

By Newton's law,
Mg - T = M*a ........(1)

a: acceleration

Because pulley is frictionless, therefore,
T = T' ; alpha' = 0 (pulley will not rotate)

For cylinder : mass m = 100 lb, theta = 50 deg, friction ...

Solution Summary

The dynamics for speed, velocity, and angles are examined. The answers are given step-by-step equationally.

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