What is an overdrive pulley?

Physics of a pulley and bucket!?

• Imagine a frictionless pulley (a solid cylinder) of unknown mass M and radius r = 0.200 m which is used to draw water from a well. A bucket of mass m = 1.50 kg is attached to a massless cord wrapped around the pulley. The bucket starts from rest at the top of the well and falls for t = 3.00 s before hitting the water h = 8.19 m below the top of the well. (a) What is the linear acceleration of the falling bucket? (b) What is the angular acceleration of the pulley? (c) What is the tension in the cord? (d) What is the torque that is applied to the pulley due to the cord? (use the tension from the previous question) (e) Using the torque and the angular acceleration, find the moment of inertia of the pulley. (f) Using the moment of inertia, find the mass of the pulley. (g) What is the change in the potential energy of the bucket? (h) What is the velocity of the bucket when it hits the water? (use the previously calculated linear acceleration) (i) What is the angular velocity of the pulley when the bucket hits the water? (use the relation between the angular velocity and the linear velocity) (j) What is the angular momentum of the pulley when the bucket hits the water? (use the moment of inertia and the angular velocity of the pulley) (k) What is the kinetic energy of the bucket when it hits the water? (l) What is the kinetic energy of the pulley when the bucket hits the water? (m) What is the total kinetic energy of the system when the bucket hits the water? (n) Is the total kinetic energy smaller, equal to (within +/- 1%), or larger than the potential energy change?

• Answer:

  Assuming that the by "frictionless" the question means no friction only on the axle of the cylinder and present elsewhere. We will right the linear and angular motions. For the bucket, Mg - T = Ma (1.5)(9.8) - T = 1.5(a) ...(1) For pulley, T x R = Iα = (MR²/2)α ...(2) a = αR ...(3) This relation comes from the fact that as the pulley rotates, it loosens rope and rope is used by the bucket to go down. Its called string constraint. Substituting αR in (2) TR = (MR/2)a T = Ma/2 Substitue T in (1) Mg - Ma/2 = Ma Mg = 3Ma/2 a = 2g/3 = 6.533 m/s² α = a/R = 6.533 / 0.2 = 32.66 rad / sec² T = Ma/2 = (1.5)(6.533)/2 = 4.9 N Torque = T x R = 4.9 x 0.2 = 0.98 N-m I = Torque/α = 0.98 / 32.66 = 0.03 kg m² M = 2I/r² = 2 x 0.03 / 0.04 = 1.5 kg Change in potential energy = Mgh where h is descended height = (1.5)(9.8)(8.19) = 120.4 J v = u + at = 0 + (6.533)3 = 19.6 m/sec Just as a = αR, v = ωR. So we can either find the velocity of the bucket when it hits the water and use this relation or directly calculate ω using angular acceleration. I am gonna use the former one. v = 19.6 ω = v/r = 19.6 / 0.2 = 98 rad/sec Angular momentum = Iω = 0.03 x 98 = 2.93 kg m^2 / sec Kinetic energy = (1/2)mv² = (1/2)(1.5)(19.6)² = 288.12 J Rotationl Kinetic energy = (1/2)Iω² = 144.06 J Total = 288.12 + 144.06 = 432.18 J The total kinetic energy is coming out to be greater than potential energy. (I have no idea why this is coming. It should have been equal)