Answer:A. To find the tension in the rope while the bucket is falling, we need to use the conservation of energy. At the top of the well, the bucket has potential energy mgh = (14.3 kg)(9.81 m/s^2)(10.5 m) = 1479 J. This potential energy is converted to kinetic energy as the bucket falls. At the bottom of the well, the bucket has only kinetic energy, given by KE = (1/2)mv^2, where v is the speed at which the bucket strikes the water. Since there is no work done by non-conservative forces like friction, the total mechanical energy is conserved. Therefore:
mgh = (1/2)mv^2 + (1/2)Iω^2,
where I is the moment of inertia of the cylinder and ω is its angular velocity. We know that the cylinder is a solid cylinder, so I = (1/2)MR^2, where M is the mass of the cylinder and R is its radius. We also know that the cylinder is rolling without slipping, so v = Rω. Substituting these expressions into the conservation of energy equation and solving for the tension T, we get:
T = mgh / (R(1 + m/M))
Plugging in the numbers, we get:
T = (14.3 kg)(9.81 m/s^2)(10.5 m) / (0.130 m(1 + 14.3 kg / 12.5 kg)) = 137 N
Therefore, the tension in the rope while the bucket is falling is 137 N.
B. To find the speed at which the bucket strikes the water, we can use the conservation of energy equation derived in part A. Solving for v, we get:
v = sqrt(2gh(M+m) / (mM + (1/2)m^2))
Plugging in the numbers, we get:
v = sqrt(2(9.81 m/s^2)(10.5 m)(12.5 kg + 14.3 kg) / ((14.3 kg)(12.5 kg) + (1/2)(14.3 kg)^2)) = 9.38 m/s
Therefore, the speed at which the bucket strikes the water is 9.38 m/s.
C. To find the time of fall, we can use the kinematic equation:
y = 1/2gt^2,
where y is the distance fallen, g is the acceleration due to gravity, and t is the time of fall. Solving for t, we get:
t = sqrt(2y/g)
Plugging in the numbers, we get:
t = sqrt(2(10.5 m)/(9.81 m/s^2)) = 1.47 s
Therefore, the time of fall is 1.47 s.
D. While the bucket is falling, the cylinder is rotating about its center of mass, which is also the axis of rotation. Since there is no net torque about this axis, the cylinder is not accelerating rotationally. Therefore, the force exerted on the cylinder by the axle is zero.
Step-by-step explanation: