In the figure below, the hanging object has a mass of m 1 = 0.410 kg; the slidin
ID: 1395537 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.410 kg; the sliding block has a mass of m2 = 0.765 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
a) work done = Change in K.E ----------------------------1
Change in K.E of system ( block m1+ block m2+ pulley)
= ( 1/2)* m1* ( Vf2 - Vi2 ) + ( 1/2)* m2*( Vf2- Vi2 ) + ( 1/2)* I * ( wf2 - wi2 )
= ( 1/2) * ( Vf2 - Vi2 ) ( m1+m2) + ( 1/2) * I * ( wf2 - wi2 ) ---------------------A
I ( moment of interia) = ( 1/2) * M * ( R12 + R22 )
V = wr, w = V/r , so wf = Vf / R2 , wi = Vi / R2
by putting all the values in eq A,
change in K.E = ( 1/2)* ( Vf2 - Vi2 ) * ( m1+m2) + ( 1/2) * ( 1/2 M * ( R12 + R22 ) ) * ( Vf2 - Vi2 / R22 )
net work done ( due to gravitational force and due to friction force on m2) :
= m1*g*d - uk *m2*g*d ( W= force*displacement)
from eq 1:
(1/2)* ( Vf2 - Vi2 ) ( m1+m2) + ( 1/4) * M * ( R12 + R22 ) * ( Vf2 - Vi2 / R22 ) = m1*g*d - uk * m2*g*d
bu putting all the values and sloving it
Vf = ( velocity after 0.700 m distance) = 1.665 m/s
b) angular speed , wf = Vf / R2 ( V = wr)
= 1.6 / 0.030
= 55.5 rad/s