A block of mass 0.72 kg is suspended by a string which is wrapped so that it is
ID: 1325619 • Letter: A
Question
A block of mass 0.72 kg is suspended by a string which is wrapped so that it is at a radius of 0.054 m from the center of a pulley. The moment of inertia of the pulley is 4.20
A block of mass 0.72 kg is suspended by a string which is wrapped so that it is at a radius of 0.054 m from the center of a pulley. The moment of inertia of the pulley is 4.20 x10^-3 kgm^2. There is friction as the pulley turns. The block starts from rest, and its speed after it has traveled downwards a distance of D= 0.83 m, is 1.805 m/s. Calculate the amount of energy dissipated up to that point.Explanation / Answer
the kinetic energy, KE = 1/2 mv^2 andthe rotational KE = 1/2 I w^2
where I is moment of inertia and w is angular velocity
the linear and angular velocity are related via v = w r so w = v/r,
w=1.805 m/s / 0.054 m = 33.426 rad/s
According to conservation of energy,
the change in PE of the block goes into the KE of the block, KE of the pulley, and work lost to friction
? mgh = 1/2 mv^2 + 1/2 I w^2 + work dissipated by friction
the work dissipated = mgh -1/2 mv^2 + 1/2 I w^2
W = 5.8565 -1/2*0.72*3.258+1/2*0.0042*1117.3
= 7.03 J
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