Chapter 6, Problem 90GP

To determine

a. The speed of the ball at the lowest point of its path.

b. The speed of the ball at the top of the given circular path about the peg.

Answer to Problem 90GP

Solution:

a. The speed of the ball at the lowest point is $v=\sqrt{2gl}$ .

b. The speed of the ball at the top of the given circular path is $v_1=\sqrt{1.2gl}$ .

Explanation of Solution

Given:

The length of the horizontal cord is $l$ .

The peg is at a distance of $h$ below the point of attachment of the cord.

This height is $h=0.80l$ .

Formula used:

The speed of the ball at the bottom of its path is calculated from the energy conservation equation. The kinetic energy of the ball is equal to its potential energy at the given lowest point.

Thus, $\frac{1}{2}m{v}^2=mgl$

Once the value of v is known, at the top of the given circular path, the speed of the ball can be calculated.

Calculations:

Part a

If the speed at the lowest point of the path is $v$ , then

  $\begin{array}{l}\frac{1}{2}m{v}^2=mgl\\ v=\sqrt{2gl}\end{array}$

Part b

The radius of the circular path about the peg is:

  $\begin{array}{l}=l-h\\ =l-0.80l\\ =0.20l\end{array}$

Therefore, its diameter is:

  $\begin{array}{l}=2\times 0.20l\\ =0.40l\end{array}$

Consider the energy conservation about this diameter of the circular path.

If the speed is $v_1$

at the top path:

  $\begin{array}{l}\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_1{}^2+mg\times 0.40l\\ \frac{1}{2}m\times 2gl=\frac{1}{2}m{v}_1{}^2+0.40mgl\\ gl=\frac{v_1{}^2}{2}+0.40gl\\ \frac{v_1{}^2}{2}=0.60gl\\ v_1=\sqrt{1.2gl}\end{array}$

Conclusion:

a. The speed of the ball at the lowest point is $v=\sqrt{2gl}$ .

b. The speed of the ball at the top of the given circular path is $v_1=\sqrt{1.2gl}$ .