CP? A 15.0-kg bucket of water is suspended by a very light rope wrapped around a solid uniform cylinder 0.300 m in diameter with mass 12.0 kg. The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 10.0 m to the water. (a) What is the tension in the rope while the bucket is falling? (b) With what speed does the bucket strike the water? (c) What is the time of fall? (d) While the bucket is falling, what is the force exerted on the cylinder by the axle?

Solution 16E Step 1 of 7: In the given problem,a M=15 kg bucket of water is suspended by a very light rope wrapped around a cylindrical wheel on the well with mass m = 12Kg with cylinder diameter d= 0.3 m falls to the distance h= 10 m. The tension in the rope which is hanging bucket of water as shown in the figure below, Using FBD for the bucket Mg-T = Ma T = Mg - Ma T= M(g-a)...........1 Step 2 of 7: Moment of inertia I for the cylinder, mr2 I= 2 Torque, = Tr = I Using I= mr2 = I = mr 2 2 Using T= M(g-a) = Tr = M(g a).r Equating above two equations, 2 M(g a).r = mr 2 mr M(g a) = 2 Using r=a M(g a) = ma 2 Step 3 of 7: Solving for acceleration, 2 Mg a = (m+2M) Substituting M=15 kg, m=12 kg and g=9.8m/s 2 2 a = 2 (15 kg)(9.8m/s ) (12 kg+2(15 kg)) 2 a= 7 m/s Step 4 of 7: (a) What is the tension in the rope while the bucket is falling Using equation equation 1, T= M(g-a) Substituting M=15 kg, a= 7 m/s and g=9.8m/s 2 2 T=15 (9.8-7) kg m/s T=42 N Therefore the tension in the rope while the bucket is falling is 42 N.