Question

A ball whirls around in a vertical circle at the end of a string. The other end of the string is fixed at the center of the circle. Assume that the total energy of the ball-Earth system remains constant. (a) What is the tension in the string at the bottom? (Use the following as necessary: m for mass of the ball, g for gravitational acceleration, vb for velocity at the bottom, and R for radius of the circle.) Tb =

Answer 1

Answer:

$T_b=m(g+\frac{v_b^2}{R})$

Explanation:

At the bottom the tension would be upwards and the weight downwards, their difference being the centripetal force. Taking the upwards direction as positive we then have:

$T_b-mg=F_{cp}=ma_{cp}=m\frac{v_b^2}{R}$

where we have used the equation for centripetal acceleration. Thus we have:

$T_b=m(g+\frac{v_b^2}{R})$