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?
a. 11.3 ft/s
b. 13.8 ft/s
c. 14.7 ft/s
d. 12.6 ft/s
e. 10.3 ft/s
Please see attached for full questions.
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)
Because pulley is frictionless, therefore,
T = T' ; alpha' = 0 (pulley will not rotate)
For cylinder : mass m = 100 lb, theta = 50 deg, friction ...
The answers are given step-by-step equationally.