A pendulum consists of a mass M hanging at the bottom end of a massless rod of length l which has a frictionless pivot at its top end. A mass m, moving as shown in Fig. 7–35 with velocity v, impacts M and becomes embedded. What is the smallest value of v sufficient to cause the pendulum (with embedded mass m) to swing clear over the top of its arc?

FIGURE 7-35Problem 42.

First, an inelastic collision occurs between two masses at the bottom of the arc; then, the kinetic energy of the masses is converted into potential energy at the top of the arc.

The mass of the pendulum is M.

The magnitude of the small mass is m.

The initial speed of mass m is v.

The initial speed of mass M is zero as it is at rest.

The length of the rod is l.

Let \(v'\) be the speed of the combined mass after the collision.

Here, since the small mass m is embedded with mass M after the collision; the collision is inelastic. Therefore, there is a loss of kinetic energy during the collision.

Now, the total momentum of the two masses before the collision is:

\(\begin{array}{c}{P_{\rm{b}}} = \left( {mv + \left( {M \times 0} \right)} \right)\\ = mv\end{array}\)

The total momentum of the masses after the collision is \(\left( {m + M} \right)v'\).

Now, using the momentum conservation, you get:

\(\begin{array}{c}\left( {m + M} \right)v' = mv\\v' = \frac{m}{{m + M}}v\end{array}\)

The combined mass will reach the top of the arc. The total mechanical energy at the bottom will be equal to the total mechanical energy at the top.

At the top, the combined mass rests as it has to reach that top point with the minimum value of v.

Let the reference height for the gravitational potential energy be at the bottom of the arc.

Now, the potential energy of the masses at the bottom of the arc is zero as they are at the reference ground, and the kinetic energy is \(\frac{1}{2}\left( {m + M} \right){\left( {v'} \right)^2}\). Therefore, the total mechanical energy of the masses at the bottom of the arc is:

\(\begin{array}{c}{E_{\rm{B}}} = \frac{1}{2}\left( {m + M} \right){\left( {v'} \right)^2} + 0\\ = \frac{1}{2}\left( {m + M} \right){\left( {\frac{m}{{m + M}}v} \right)^2}\\ = \frac{1}{2}\frac{{{m^2}}}{{m + M}}{v^2}\end{array}\)

Now, the potential energy of the masses at the top of the loop is \(2\left( {m + M} \right)gl\) , and the kinetic energy is zero. The total mechanical energy of the masses at the top of the arc is:

\(\begin{array}{c}{E_{\rm{T}}} = 0 + 2\left( {m + M} \right)gl\\ = 2\left( {m + M} \right)gl\end{array}\)

Now, using energy conservation, you get:

\(\begin{array}{c}{E_{\rm{B}}} = {E_{\rm{T}}}\\\frac{1}{2}\frac{{{m^2}}}{{m + M}}{v^2} = 2\left( {m + M} \right)gl\\{v^2} = 4{\left( {\frac{{m + M}}{m}} \right)^2}gl\\v = 2\frac{{m + M}}{m}\sqrt {gl} \end{array}\)

Hence, the smallest value of v that is sufficient to cause the pendulum to swing over the top of the given arc is \(2\frac{{m + M}}{m}\sqrt {gl} \).