You pull downward with a force of 29 on a rope that passes over a disk-shaped pu

Question

You pull downward with a force of 29 on a rope that passes over a disk-shaped pulley of mass 1.4 and radius 0.075 . The other end of the rope is attached to a 0.65 mass. Find the linear acceleration of the 0.65 mass.

Explanation / Answer

In this problem where the pulley’s mass is not negligible and hence it has to be considered in our calculations, you have to remember that the tension in the rope in one side of the pulley is not anymore the same tension on the other side of the pulley. The same tension on both sides of the pulley only happens in theory if we consider the pulley’s mass to be negligible. You need to start the analysis with free body diagrams for both sides of the pulley. In one side (say the left side) we have the mass m (0.64Kg) hanging from the rope. This mass will move up due to the force F applied to the rope on the right side of the pulley. Free Body Diagram on the left: m x a = T1 – (m x g) => T1 = m x a + m x g Free Body Diagram on the right: F = T2 Now we need to find the equations of the pulley relating both T1 and T2: The magnitude of T2 on the right side of the pulley is larger than T1 and for this reason the mass m will move up. (T2 – T1) x R = I x alpha Where R is the pulley’s radii, I is the moment of Inertia of the pulley (I = MxR^2/2) and alpha is angular acceleration. Now we need to find a relation between the angular and linear acceleration, which is: a = alpha x R (T2 – T1) x R = I x a / R When you substitute T1 = m.a + m.g and F = T2 in (T2 – T1) x R = I x a / R and solve for a you get: a = (T2 – m.g) / (m +M/2) where M is pulley’s mass, m is the hanging mass, T2 is the force applied to the rope. a = (F – m.g) / (m +M/2) Numerically you get: a = 17.52 m/s2 If the pulley’s mass were negligible, the acceleration would be 33.93 m/s2.

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