Question

What is the minimum speed the rock must have at the top of the circle if it is to always stay in contact with the bottom of the bucket?

Answer 1

The bucket, and so also the rock, is moving in a circular path. In order to keep the rock at the bottom of the rock, you need to swing the bucket fastenough that the force of gravity at the highest point on the circle is equal to the centripetal force experienced by the rock. This is because when therock in the bucket is at its highest point, the only force acting on it is gravity and while moving in a circle it has a net acceleration directed towardsthe center of the circle, so if the force of gravity is just equal to the centripetal force, then that force is causing the rock to remain in a circularpath, but there is no longer a force to pull it away from the bucket, in a sense, the force of gravity is "used up" keeping the rock in a circular path.The law of inertia states that the rock will not fall out of the bucket because there is no force to make it do so. The equations for this are asfollowos

Forces on rock = net force
Fg = Fc
mg = ma
in a circular path a=v^2 / r where r is the tangential velocity
mg = m*(v^2)/r
divide the m from both sides and solve for v
v=sqrt(g*r)
you will notice that the speed you need to spin the bucket is independent of the weight of the rock, so whether the rock weighs 500g or 500 kg the speedyou need to spin it is the same. The tension you feel in your arm will increase as the weight increases, but the necessary speed will be unchanged.

Answer 2

You hold a bucket in one hand, and in the bucket is a 500g rock. You swing the bucket so the rock moves in a vertical circle 2.2m in diameter. (this wasthe setup to the question) ^answer is correct, so it would be v= sqrt(9.8*1.1) v=3.28 which is 3.3 m/s with correct sigfigs