- #1

Kennedy

- 70

- 2

## Homework Statement

A mass m, at one end of a string of length L, rotates in a vertical circle just fast enough to prevent the string from going slack at the top of the circle. Assuming mechanical energy is conserved, the tension in the string at the bottom of the circle is: a) 6 mg b) mg + 3mg/L c) 5 mg d) 5gL e) 4 mgL

## Homework Equations

a = v^2/r

KEi + PEi = KEf + KEi

## The Attempt at a Solution

I understand that the centripetal acceleration at the top of the circle is 9.8 m/s^2, because then there is no tension in the string because the centripetal force all comes from the weight of the object. I understand that the PE at the top of the circle is 2L(g)(m), considering that the bottom of the circle is 0 PE. Now, since energy is conserved, the speed of the mass at the bottom of the circle is found by using the fact that the negative of the change in potential energy is equal to the kinetic energy. At the bottom of the circle, all of the potential energy is converted into kinetic energy. so, 2(L)(g)(m) = 1/2(m)(v^2). This yields v = (4Lg)^(1/2). So, to find the centripetal force at the bottom of the motion (otherwise known as the tension in the string) a = 4Lg/L = 4g, then Fc = m(4g) = 4mg, but clearly this isn't an option, but the answer that makes the most sense to me.