Question

A light, rigid rod is 77.0 cm long. Its top end is pivoted on a frictionless, horizontal axle. The rod hangs straight down at rest with a small, massive ball attached to its bottom end. You strike the ball suddenly giving it a horizontal velocity so that it swings around in a full circle. What minimum speed at the bottom is required to make the ball go over the top of the circle?

Answer 1

for the ball to move in a circle,
\[v=\sqrt{gr}\]

Answer 2

in this case only conservative forces act. so apply conservation of mechanical energy. here you will have to consider both rotational and translational kinetic energies.

Answer 3

Wouldn't that basically be Newton's Second Law=Centripetal acceleration?
I need my answer using energy...

Answer 4

@chand I don't know anything about rotational energy, only centripetal acceleration.

Answer 5

oh. sorry. here the rod is light. so no rotational energy. only translational energy.
\[\Delta PE = KE\]

Answer 6

@chand I do know about translational energy though (I think) It's basically if there is only conservative forces acting on an isolated system, the sum of kinetic energy and potential energy (gravitational and elastic)=0 right?

Answer 7

not exactly. its: decrease in potential energy = gain in kinetic energy of the body.
in this case decrease in kinetic energy = gain in potential energy.

Answer 8

Isn't that the sum of kinetic and potential=0? I just do some algebraic manipulation.

Answer 9

\[\Delta PE + \Delta KE \] is not zero in this case, though it is constant.

Answer 10

yes. its a constant. thats why we use the conservation of energy

Answer 11

If it was zero you would get escape velocity. Ad notam))

Answer 12

Ok, so what direction is the circle going in? Is it a vertical circle or a horizontal one?

Answer 13

its a vertical circle.

Answer 14

But it has a horizontal velocity...

Answer 15

For the top of the circle we don't need any kinetic energy so there we get for KE
KE = 0
But
\[PE = KE _{at start}=KE _{\max}\]