In the figure below, the hanging object has a mass of m 1 = 0.495 kg; the slidin
ID: 1449897 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.495 kg; the sliding block has a mass of m2 = 0.780 kg; and the pulley is a hollow cylinder with a mass of M = 0.350 kg, an inner radius of R1 = 0.020 0 m, and an outer radius of R2 = 0.030 0 m. Assume the mass of the spokes is negligible. The coefficient of kinetic friction between the block and the horizontal surface is ?k = 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of vi = 0.820 m/s toward the pulley when it passes a reference point on the table.
(a) Use energy methods to predict its speed after it has moved to a second point, 0.700 m away.
(b) Find the angular speed of the pulley at the same moment.
Explanation / Answer
Here ,
m1 = 0.495 Kg
m2 = 0.78 Kg
M = 0.350 Kg
R1 = 0.020 m
R2 = 0.030 m
uk = 0.250
vi = 0.820 m/s
for d = 0.70 m/s
I = 0.5 * M * (R1^2 + R2^2)
I = 0.5 * 0.350 * (0.020^2 + 0.030^2)
I = 2.275 *10^-4 Kg.m^2
a) Using conservation of energy
0.5 * m1 * v^2 + 0.5 * m2 * v^2 + 0.5 * I * (v/R2)^2 = 0.5 * 0.495 * 0.820^2 + 0.5 * 0.495 * 0.820^2 + 0.495 * 9.8 * 0.70 - 0.250 * 0.780 * 9.8 * 0.8
0.5 * 0.495 * v^2 + 0.5 * 0.78 * v^2 + 0.5 * 2.275 *10^-4 * (v/0.20)^2 = 0.5 * 0.495 * 0.820^2 + 0.5 * 0.495 * 0.820^2 + 0.495 * 9.8 * 0.70 - 0.250 * 0.780 * 9.8 * 0.8
solving for v
v = 1.853 m/s
the speed after moving 0.70 m away is 1.853 m/s
b)
angular speed = v/R2
angular speed = 1.853/(0.03)
angular speed = 61.7 rad/s
the angular speed of the pulley is 61.7 rad/s