(II) A bucket of mass 2.00 kg is whirled in a vertical circle of radius 1.20 m . At the lowest point of its motion the tension in the rope supporting the bucket is 25.0 N. (a)Find the speed of the bucket (b) How fast must the bucket move at the top of the circle so that the rope does not go slack?

(a) At the lowest point of the bucket's motion, the forces acting on the bucket are the tension in the rope (T) and the weight of the bucket (mg). The net force acting on the bucket is the centripetal force (F_c), which is given by the equation F_c = T - mg. The centripetal force is also given by the equation F_c = (mv^2)/r, where m is the mass of the bucket, v is the speed of the bucket, and r is the radius of the circle. Equating the two expressions for the centripetal force, we get:<br /><br />T - mg = (mv^2)/r<br /><br />Solving for v, we get:<br /><br />v = sqrt((T - mg)r/m)<br /><br />Substituting the given values, we get:<br /><br />v = sqrt((25.0 N - (2.00 kg)(9.8 m/s^2))(1.20 m)/(2.00 kg))<br />v = 3.05 m/s<br /><br />Therefore, the speed of the bucket at the lowest point of its motion is 3.05 m/s.<br /><br />(b) At the top of the circle, the forces acting on the bucket are the tension in the rope (T) and the weight of the bucket (mg). The net force acting on the bucket is the centripetal force (F_c), which is given by the equation F_c = T + mg. The centripetal force is also given by the equation F_c = (mv^2)/r, where m is the mass of the bucket, v is the speed of the bucket, and r is the radius of the circle. Equating the two expressions for the centripetal force, we get:<br /><br />T + mg = (mv^2)/r<br /><br />Solving for v, we get:<br /><br />v = sqrt((T + mg)r/m)<br /><br />Substituting the given values, we get:<br /><br />v = sqrt((25.0 N + (2.00 kg)(9.8 m/s^2))(1.20 m)/(2.00 kg))<br />v = 4.43 m/s<br /><br />Therefore, the bucket must move at a speed of 4.43 m/s at the top of the circle so that the rope does not go slack.