You pull downward with a force of 30 on a rope that passes over a disk-shaped pu

Question

You pull downward with a force of 30 on a rope that passes over a disk-shaped pulley of mass 1.2 and radius 0.075 . The other end of the rope is attached to a 0.68 mass. Find the linear acceleration of the 0.68 mass. Answer pleasE!! thanks! will rate!!!

Explanation / Answer

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

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