In the figure below, the hanging object has a mass of m_ 1 = 0.400 kg; the slidi
ID: 1446618 • Letter: I
Question
In the figure below, the hanging object has a mass of m_ 1 = 0.400 kg; the sliding block has a mass of m_ 2 = 0.785 kg; and the pulley is a hollow cylinder with a mass of M = 0.350 kg, an inner radius of R_ 1 = 0.020 0 m, and an outer radius of R_ 2 = 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 mew_ 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 v_ i = 0.820 m/s toward the pulley when it passes a reference point on the table. Use energy methods to predict its speed after it has moved to a second point, 0.700 m away. Find the angular speed of the pulley at the same moment.Explanation / Answer
The change in kinetic energy of the system ( both block and pulley) is equal to the net workdone on the system
KE+ KE_r = m1gh- uk m2g
1/2 ( m1+ m2) ( vf^2- vi^2) + 1/2 ( wf^2 - wi^2) = m1gh- uk m2g
1/2 ( m1+ m2) ( vf^2- vi^2) + 1/2 (* 1/2 M (R1^2 + R2^2) ( vf^2- vi^2/R2^2)= m1gh- uk m2g
solving for vf
vf = sqrt vi^2 + (m1gh- uk m2g/ 1/2( m1+m2) + 1/2 M( 1+ R1^2/R2^2))
=sqrt ( 0.820)^2 + (0.4(9.8)(0.7)- 0.250(0.785)(9.8)/ 1/2(0.4+0.785) + 1/2 (0.350( 1+(0.0200)^2/(0.0300)^2
=sqrt ( 0.820)^2 +((0.4(9.8)(0.7)- 0.250(0.785)(9.8)/0.7675
=1.319 m/s
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(b)
angular speed of the pulley is
w = vf/R2 = 1.319/0.030 = 43.99 rad/s