A small mass m attached to the end of a string revolves in a circle on a frictionless table top. The other end of the string passes through a hole in the table (Fig. 8–62). Initially, the mass revolves with a speed\({v_1} = 2.4\;{\rm{m/s}}\)in a circle of radius\({r_1} = 0.80\;{\rm{m}}\). The string is then pulled slowly through the hole so that the radius is reduced to\({r_2} = 0.48\;{\rm{m}}\). What is the speed,\({v_2}\), of the mass now?

The given mass is m.

The initial velocity of the mass is\({v_1} = 2.4\;{\rm{m/s}}\).

The initial radius of the circular motion is\({r_1} = 0.80\;{\rm{m}}\).

The radius of the circle traced after pulling the string is \({r_2} = 0.48\;{\rm{m}}\).

Angular momentum is the rotational equivalent of the linear momentum of a body.It is expressed as the product of the moment of inertia and the angular velocity.

\(L = I\omega \)

Angular momentum can also be written as the product of linear momentum and the radius of the circular motion.

\(L = mvr\)

The string is pulled at an angle of\({90^{\rm{o}}}\)to the velocity of the mass. As there is no net external torque acting on the system, the total angular momentum remains conserved.

\(\begin{aligned}{c}{I_1}{\omega _1} = {I_2}{\omega _2}\\m{v_1}{r_1} = m{v_2}{r_2}\\{v_1}{r_1} = {v_2}{r_2}\\{v_2} = \frac{{{v_1}{r_1}}}{{{r_2}}}\end{aligned}\)

Substituting the known numerical values in the above expression, you get:

\(\begin{aligned}{c}{v_2} = \frac{{\left( {2.4\;{\rm{m/s}}} \right)\left( {0.80\;{\rm{m}}} \right)}}{{0.48\;{\rm{m}}}}\\ = 4\;{\rm{m/s}}\end{aligned}\)

Thus, the speed \({v_2}\) of the mass is \(4.0\;{\rm{m/s}}\).