Chapter 12, Problem 49A

A 0.60-kg puck revolves at 2.4 m/s at the end of a 0.90-m string on a frictionless air table. Show that when the string is shortened to 0.60 m the speed of the puck will be 3.6 m/s.

To determine

To prove: The speed of the considered puck will be $3.6\text{m/s}$ when the string is shortened to $0.60\text{m}$ .

Explanation of Solution

Given:

The mass of the puck, $m=0.60\text{kg}$

The speed of the puck, $v_1=2.4\text{m/s}$

The initial length of the string, $r_1=0.90\text{m}$

Formula used:

The angular momentum of mass $m$ moving in a circle of radius $r$ with a speed $v$ is given by,

  $L=mvr$

And, when no external torque acts on a system, then the angular momentum of the system remains conserved.

Proof:

Since no external torque is exerted on the puck, hence the angular momentum of the puck-string remains conserved.

Then, using the given values,

  $\begin{array}{c}{L}_1=L_2\\ m{v}_1{r}_1=m{v}_2{r}_2\\ v_2=\frac{v_1{r}_1}{r_2}\end{array}$

Using the given values in above,

  $\begin{array}{c}{v}_2=\frac{\left(2.4\right)\left(0.90\right)}{0.60}\\ =3.6\text{}\text{m/s}\end{array}$

Conclusion:

Hence, when the string is shortened to $0.60\text{m}$ , the speed of the considered puck will become $3.6\text{m/s}$ .