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What is acceleration?

What is the linear acceleration of the falling 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 = 5.85 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?

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    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 = 5.85 m below the top of the well. The average velocity of the bucket = 5.85m ÷ 3.00 s = 1.95 m/s The bucket accelerates at a constant rate. Initial velocity = 0 m/s. So the final velocity = 2 * 1.95 = 3.9 m/s Linear acceleration = (vf – vi) ÷ t = (3.9 – 0) ÷ 3 = 1.3 m/s^2 Angular acceleration = Linear acceleration ÷ radius = 1.3 ÷ 0.2 = 6.5 radians/s^2 This is the angular acceleration of the pulley. Weight – Tension = mass of bucket * acceleration of bucket 1.5 * 9.8 – T = 1.5 * 1.3 14.7 – T = 1.95 T = 14.7 – 1.95 = 12.75 N The torque caused by the tension of the cord cause the pulley to accelerate. Torque = T * r = 12.75 * 0.2 = 2.55 N * m and Torque = I * α = I * 6.5 I * 6.5 = 2.55 I = 2.55/6.5 and I = ½ * M * 0.2^2 = M * 0.02 M * 0.02 = 2.55/6.5 Mass of Pulley = (2.55/6.5) ÷ 0.02 PE = m * g * h = (2.55/6.5) ÷ 0.02 * 9.8 * 5.85 (h) What is the velocity of the bucket when it hits the water? (use the previously calculated linear acceleration) vf = vi + a * t = vi = 0 m/s. a = 1.3 m/s t = 3 s (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) ωf = vf ÷ r (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) angular momentum = I * ωf (k) What is the kinetic energy of the bucket when it hits the water? KE = ½ * m * vf^2 (l) What is the kinetic energy of the pulley when the bucket hits the water? ½ * m * vf^2 (m) What is the total kinetic energy of the system when the bucket hits the water? SUM

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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 = 5.85 m below the top of the well. The average velocity of the bucket = 5.85m ÷ 3.00 s = 1.95 m/s The bucket accelerates at a constant rate. Initial velocity = 0 m/s. So the final velocity = 2 * 1.95 = 3.9 m/s Linear acceleration = (vf – vi) ÷ t = (3.9 – 0) ÷ 3 = 1.3 m/s^2 Angular acceleration = Linear acceleration ÷ radius = 1.3 ÷ 0.2 = 6.5 radians/s^2 This is the angular acceleration of the pulley. Weight – Tension = mass of bucket * acceleration of bucket 1.5 * 9.8 – T = 1.5 * 1.3 14.7 – T = 1.95 T = 14.7 – 1.95 = 12.75 N The torque caused by the tension of the cord cause the pulley to accelerate. Torque = T * r = 12.75 * 0.2 = 2.55 N * m and Torque = I * α = I * 6.5 I * 6.5 = 2.55 I = 2.55/6.5 and I = ½ * M * 0.2^2 = M * 0.02 M * 0.02 = 2.55/6.5 Mass of Pulley = (2.55/6.5) ÷ 0.02 PE = m * g * h = (2.55/6.5) ÷ 0.02 * 9.8 * 5.85 (h) What is the velocity of the bucket when it hits the water? (use the previously calculated linear acceleration) vf = vi + a * t = vi = 0 m/s. a = 1.3 m/s t = 3 s (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) ωf = vf ÷ r (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) angular momentum = I * ωf (k) What is the kinetic energy of the bucket when it hits the water? KE = ½ * m * vf^2 (l) What is the kinetic energy of the pulley when the bucket hits the water? ½ * m * vf^2 (m) What is the total kinetic energy of the system when the bucket hits the water? SUM

electron...

small correction to electron1 PE=m*g*h =1.5*9.8*5.85

Rob

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