In the figure below, the hanging object has a mass of m 1 = 0.405 kg; the slidin
ID: 1396122 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.405 kg; the sliding block has a mass of m2 = 0.845 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.
m/s
(b) Find the angular speed of the pulley at the same moment.
rad/s
Explanation / Answer
here,
let the final speed of the blocks is v
final angular speed of pulley is w
for the pulley , as the rope is not slipping
w = v/R2
m1 = 0.405 kg
m2 = 0.845 kg
mass of cyclinder , m = 0.350 kg
R1 = 0.020 m
R2 = 0.030 m
moment of inertia of hollow cyclinder , I = 0.5 * m * (R1^2 + R2^2)
I = 0.5 * 0.35 * (0.20^2 + 0.30^2)
I = 0.022 kg . m^2
using conservation of energy
change in kinetic energy = - change in potential energy - work done by friction
final kinetic energy - initial kinetic energy = - change in potential energy - work done by friction
(0.5 * 0.845 * v^2 + 0.5 * 0.5 * 0.405*v^2 + 0.5 * 0.022 * (v/0.030)^2) - (0.5 * 0.845 * 0.820^2 + 0.5 * 0.5 * 0.405*0.82^2 + 0.5 * 0.022 * (0.82/0.030)^2) = 0.405 * 9.8 * 0.7 - 0.845 * 0.25 * 9.8 * 0.7
v = 0.94 m/s
speed after it has moved to a second point is 0.94 m/s
(b)
and we know
v = R2 * w
0.94 = 0.030 * w
w = 31.33 rad/s
the angular speed of the pulley at the same moment is 31.33 rad/s