In the figure below, the hanging object has a mass of m1 = 0.355 kg; the sliding
ID: 1477628 • Letter: I
Question
In the figure below, the hanging object has a mass of m1 = 0.355 kg; the sliding block has a mass of m2 = 0.860 kg; and the pulley is a hollow cylinder with a mass ofM = 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 ofvi = 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
part a:
initial kinetic energy of block m2=0.5*m2*vi^2
=0.5*0.86*0.82^2=0.2891 J
initial kinetic energy of block m1=0.5*m1*vi^2=0.11935 J
moment of inertia of the pulley=0.5*M*(R1^2+R2^2)
=0.5*0.35*(0.02^2+0.03^2)=2.275*10^(-4) kg.m^2
angular veloicty of pulley=vi/R2=0.82/0.03=27.33 rad/s
then rotational kinetic energy of the pulley=0.5*moment of inertia*angular veloicty^2
=0.5*2.275*10^(-4)*27.33^2
=0.085 J
hence total energy of the system at the reference instant=0.2891+0.11935+0.085=0.49345 J
when the block moves 0.7 m, energy is lost in work done against friction
friction force=friction coefficient*nomral force
=0.25*0.86*9.8=2.107 N
then work done against friction for moving 0.7 m =2.107*0.7=1.4749 J
as m1 will drop a height of 0.7 m , decrease in potential energy=-m1*g*0.7
=2.4353 J
hence total final kinetic energy will become 2.4353+0.49345-1.4749=1.45385 J
(decrease in potential energy mean increase in kinetic energy in order to keep total energy constant
and energy will decrease as energy will be spent against friction )
hence if final speed is v,
then 0.5*m1*v^2+0.5*m2*v^2+0.5*2.275*10^(-4)*(v/0.03)^2=1.45385
==>0.7338*v^2=1.45385
==>v=1.4075 m/s
part b:
angular speed=linear speed/outer radius=1.4075/0.03=46.9167 rad/sec