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 = 9.99 m below the top of the well.
(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?
Is the total kinetic energy smaller, equal to (within +/- 1%), or larger than the potential energy change?
2006-10-25 17:16:49 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
1) the angular velocity = v * r = 9.99/3 * .2 = 0.666 r/s
2) the angular momentum = the angular velocity * time = 0.666*3 = 1.998 r
3) ke"at the potum" = pu"at the top" = m*g*h = 1.5 * 9.8 * 9.99 = 146.853 j
4) ke of the pulley = ke of the bucket = 146.853 j " because there is no friction"
5) total ke = ke " at the putom " - ke " at the top" = 146.853- zero = 146.853j
6) total ke = - total pu
2006-10-25 23:32:19 · answer #1 · answered by mms 3 · 0⤊ 0⤋
the bucket can fall no faster than the rope can unwind so set the angular acceleration of the cylinder equal to m g r, that is the force times r, which causes the pulley to turn.
2006-10-26 06:40:33 · answer #2 · answered by meg 7 · 0⤊ 0⤋