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. Assuming the total energy of the ball– Earth system remains constant, show that the tension in the string at the bottom is greater than the tension at the top by six times the ball’s weight.

Short answer

The tension in the string at the bottom is greater than the tension at the top by six times the ball’s weight.

Step-by-step solution

Step 1:

For an isolated system, the change in the kinetic energy must be equal and opposite to the potential energy of the system. It is the direct result of the law of conservation of energy.

$\mathbf{\Delta K}\mathbf{+}\mathbf{\Delta U}\mathbf{=}\mathbf{0}$

Step 2:

Let the subscripts $\mathrm{b}$ and $\mathrm{t}$ represent the values at the bottom and thetop of the circle respectively.

The equations of motion at the top and bottom of the circle are:

${\mathrm{T}}_{\mathrm{b}}-\mathrm{mg}=\frac{{\mathrm{mv}}_{\mathrm{b}}^2}{\mathrm{R}}$

$-{\mathrm{T}}_{\mathrm{t}}-\mathrm{mg}=-\frac{{\mathrm{mv}}_{\mathrm{t}}^2}{\mathrm{R}}$

Adding these equation gives:

${\mathrm{T}}_{\mathrm{b}}={\mathrm{T}}_{\mathrm{t}}+2\mathrm{mg}+\frac{\mathrm{m}({\mathrm{v}}_{\mathrm{b}}^2-{\mathrm{v}}_{\mathrm{t}}^2)}{\mathrm{R}}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

From the law of conservation of energy, we get:

$\begin{array}{c}\mathrm{\Delta K}+\mathrm{\Delta U}=0\\ \frac{\mathrm{m}({\mathrm{v}}_{\mathrm{b}}^2-{\mathrm{v}}_{\mathrm{t}}^2)}{2}+(0-2\mathrm{mgR})=0\\ \frac{\mathrm{m}({\mathrm{v}}_{\mathrm{b}}^2-{\mathrm{v}}_{\mathrm{t}}^2)}{\mathrm{R}}=4\mathrm{mg}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)\end{array}$

By replacing equation (2) in equation (1), we get:

${\mathrm{T}}_{\mathrm{b}}={\mathrm{T}}_{\mathrm{t}}+6\mathrm{mg}$

Hence, it has been proven that the tension in the string at the bottom is greater than the tension at the top by six times the ball’s weight.