Question

A ball of mass m is attached to a string of length L. It is being swung in a vertical circle with enough speed so that the string remains taut throughout the ball's motion. Assume that the ball travels freely in this vertical circle with negligible loss of total mechanical energy. At the top and bottom of the vertical circle, the ball's speeds are vt andvb, and the corresponding tensions in the string are Tt and Tb. Tt and Tb have magnitudes Tt and find Tb-Tt, the difference between the magnitude of the tension in the string at the bottom relative to that at the top of the circle. Express the difference in tension in terms of m and g.

Answer 1

Answer:

$\frac{m}{r}(v_b^2-v_t^2)+2mg$

Explanation:

When the ball is at the bottom position of the vertical circle, the forces acting on the ball are:

- The tension in the string, $T_b$, upward

- The weight of the ball, $mg$, downward

The resultant of these forces must be equal to the centripetal force, which points upward as well (towards the center of the circle), so:

$T_b-mg=m\frac{v_b^2}{r}$ (1)

where $v_b$ is the speed of the ball at the bottom of the circle, r the radius of the circle, and m the mass of the ball.

When the ball is at the top position of the vertical circle, the weight still acts downward, however the tension in the string also acts downward, so the equation of the forces becomes:

$T_t+mg=m\frac{v_t^2}{r}$ (2)

where

$T_t$ is the tension in the string in the top position

$v_t$ is the speed of the ball in the top position

By subtracting eq.(2) from eq.(1), we find:

$T_b-mg-(T_t+mg)=m\frac{v_b^2}{r}-m\frac{v_t^2}{r}\\T_b-T_t=\frac{m}{r}(v_b^2-v_t^2)+2mg$

So, this is the difference in tension between the two positions.

Answer 2

Answer:

6mg

Explanation:

The method outlined in the hints is really the only practical way to do this problem. If done properly, finding the difference between the tensions,  Tb−Tt, can be accomplished fairly simply and elegantly.