Thread: Motion in a vertical circle

Hello this is a question from A2 level mechanics, am struggling to get the book answer. Here goes:

A bead, of mass m is threaded onto a smooth circular ring, of radius r, which is fixed in a vertical plane. The bead is moving on the wire so that its speed at the lowest point of its path is four times its speed, v, at the highest point.

Find v in terms of r and g.

My answer:
The usual technique is to work in terms of energy thus,

Loss in kinetic energy = gain in potential

1/2mv^2 - 1/2m(v/4)^2 = mgr

1/2m(v^2 - (v^2)/16) = mgr

(15/16)v^2 = 2gr

v^2 = 32gr/15

v = root(32gr/15)

But the book answer is v = root(4gr/15)

Any help offered is greatly appreciated.

Originally Posted by gtbiyb

Hello this is a question from A2 level mechanics, am struggling to get the book answer. Here goes:

A bead, of mass m is threaded onto a smooth circular ring, of radius r, which is fixed in a vertical plane. The bead is moving on the wire so that its speed at the lowest point of its path is four times its speed, v, at the highest point.

Find v in terms of r and g.

My answer:
The usual technique is to work in terms of energy thus,

Loss in kinetic energy = gain in potential

1/2mv^2 - 1/2m(v/4)^2 = mgr

1/2m(v^2 - (v^2)/16) = mgr

(15/16)v^2 = 2gr

v^2 = 32gr/15

v = root(32gr/15)

But the book answer is v = root(4gr/15)

Any help offered is greatly appreciated.

HI

By the energy principles ,







Therefore , which agrees with your answer .

you should take the LOSS in kinetic energy , not kinetic energy , and also the height here is 2r , the diameter of the vertical circle . Hope its clear .

Thank you,

Yeah change in height is 2r, silly mistake. I can see i need the change in kinetic energy but i formed the equation as the velocity at the top is 1/4 that of the bottom. Thought that would result in same answer but it doesnt! Nevermind thanks again.