Chapter 11.2, Problem 20PP

To determine

The linear speed of the ball after the pinball leaves the spring.

Answer to Problem 20PP

  $0.845\text{m/s}$

Explanation of Solution

Given:

The mass of the ball, $m=0.08\text{kg}$

The force exerted on the ball by the spring, $F=2\text{N}$

The distance upto which the force is exerted, $d=5\text{cm}=0.05\text{m}$

Moment of inertia of the ball, $I=\frac{5}{2}m{r}^2$

Formula used:

Work energy theorem,

  $W=K{E}_i-K{E}_f$

Where W is the work done, and $KE$ is the kinetic energy.

The work done is given by,

  $W=Fd\cos \theta$

Where $F$ is the force done, $d$ is the distance, and $\theta$ is the angle between the distance and force.

The linear kinetic energy of an object of mass $m$ travelling at a speed of $v$ is given by,

  $K{E}_l=\frac{1}{2}m{v}^2$

And, the rotational kinetic energy of an object of moment of Inertia I , and travelling with angular velocity $w$ is given by,

  $K{E}_r=\frac{1}{2}I{\omega }^2$

Calculation:

Let the linear speed of the pinball be v .

Now, the rotational kinetic energy of the pinball will be,

  $\begin{array}{c}K{E}_r=\frac{1}{2}I{\omega }^2\\ =\frac{1}{2}\left(\frac{5}{2}m{r}^2\right){\omega }^2\end{array}$

It is known as, $\omega =\frac{v}{r}$ . Then, the rotational kinetic energy of the pinball will be,

  $\begin{array}{c}K{E}_r=\frac{1}{2}I{\omega }^2\\ =\frac{1}{2}\left(\frac{5}{2}m{r}^2\right){\left(\frac{v}{r}\right)}^2\\ =\frac{5}{4}m{v}^2\end{array}$

Using work energy theorem and the formula of work done, then,

  $\begin{array}{c}W=K{E}_f-K{E}_i\\ Fd\cos 0°=\left(K{E}_l+K{E}_r\right)-\left(0\right)\\ Fd=\left(\frac{1}{2}m{v}^2+\frac{5}{4}m{v}^2\right)-\left(0\right)\\ Fd=\frac{7}{4}m{v}^2\end{array}$

Using given values in above,

  $\begin{array}{c}Fd=\frac{7}{4}m{v}^2\\ \left(2\right)\left(0.05\right)=\frac{7}{4}\left(0.08\right)v^2\\ v=0.845\text{m/s}\end{array}$

Conclusion:

Thus, the linear speed of the ball after the pinball leaves the spring is $0.845\text{m/s}$ .