Consider two objects with m 1 > m 2 connected by a light string that passes over
ID: 2262224 • Letter: C
Question
Consider two objects with m1 > m2 connected by a light string that passes over a pulley having a moment of inertia of I about its axis of rotation as in the figure below. The string does not slip on the pulley or stretch. The pulley turns without friction. The two objects are released from rest separated by a vertical distance 2h. (Use any variable or symbol stated above along with the following as necessary: g and R.)
(a) Use the principle of conservation of energy to find the translational speeds of the masses as they pass each other.
(b) Find the angular speed of the pulley at this time.
Explanation / Answer
it forces you to assum that m1 is initially higher than m2 otherwis they don't pass each other. No matter.
Start with the potential energy of the system. Assume the rope has a length L = pi*R + 2h where R = radiusof the pulley. Then m1 is initially right up against the pulley and m2 is 2h unit down from that. Call the initial position of m2 the point where teh gravitational kinetic energy is zero. So initially we have a total energy that is all potential energy, given by:
Ei = Ui =m1*g*2h =2*m1*g*h
Now release the masses and m1 falls causing m2 to rise. Since teh rope does not stretch or slip, both masses have teh same speed v, at any point in time. The pulley also gain kinetic energy from this since it rotates and has a moment of interia. So at some point y in the vertical direction we can write teh potential energy as:
U(y) = m1*g*(2h - y) + m2*g*y
The kinetic energy at any point y is
T(y) = 1/2*m1*v^2 + 1/2*m2*v^2 +1/2 I*w^2
where w = angular speed of the pulley. Now since the rope does not stretch or slip, the rope is moving at speed v. Since the rope is in tangential contact with the pulley, the linear speed v and angualr speed v are related:
w = v/R so
T(y) = 1/2*m1*v^2 + 1/2*m2*v^2 +1/2 I*(v^2/R^2)
Now U(y) + T(y) = Ui by conservation of energy so
2*m1*g*h = 1/2*(m1 + m2 *I/R^2)*v^2 + m1*g*(2h - y) + m2*g*y
Now when they pass each other, they are both a distance h away from the zero potential energy point so set y = h
2*m1*g*h = 1/2*(m1 + m2 *I/R^2)*v^2 + m1*g*h + m2*g*h
And solve for v
(2*m1 - 2*m2)*g*h = (m1 + m2 *I/R^2)*v^2 ---> v = sqrt(((2*m1 - 2*m2)*g*h)/(m1 + m2 *I/R^2))