Question

A bead slides without friction around a loopthe-loop. The bead is released from a height of 18.3 m from the bottom of the loop-theloop which has a radius 6 m. The acceleration of gravity is 9.8 m/s 2 . What is its speed at point A ? Answer in units of m/s

Answer 1

Conservation of energy might help, considering that kinetic energy is \(K = \dfrac{mv^2}{2}\) and potential energy is \(U = mgh\).

Answer 2

I don't understand? :l I'm extremely bad at physics

Answer 3

Where's point A at? Note that since friction is not a factor, we only need to know the height of point A.

To expand upon yak's insight, the law of the conservation of energy tells us that the total energy of the system must not change. In equation form, we have:

(initial kinetic energy) + (initial potential energy) = (final kinetic energy) + (final potential energy)

The initial energy, in this case, is just the energy from, I assume, dropping the bead from a certain height -- there is no initial velocity. So, what's next? I'll start it:

(18.3)(mass of bead)(9.8) = (final kinetic energy) + (final potential energy)

So, we seem to need to know a couple more things to figure out the final speed: (1) mass of bead, (2) final potential energy.