Q7.14 Bouncing a ball (S)

A ball of mass m rests on a bat which executes vertical simple harmonic motion of frequency ω and amplitude xm.
  1. Derive an expression for the force exerted by the bat on the ball.
  2. Assuming an oscillation period of 1 s, determine the minimum amplitude of oscillation at which the bat and ball will separate.
  3. Assuming an amplitude of 5 cm determine the maximum frequency for which the bat and ball remain in contact.
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Hint

You need to reverse the usual argument. Here you are told that the ball does SHM. So you know what its acceleration has to be when the bat (and ball) are displaced by x. So you can infer what the net force is acting on it. What are the forces (two of them!) that contribute? What can you say about one of those forces when the bat and ball are just about to part company?
 

Solution

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Solution

As long as the ball remains in contact with the bat it will exhibit SHM.

Let x denote the upward displacement of the ball from the position it occupies at the middle of the SHM cycle. We will consider separately (a) the upward acceleration of the ball and (b) the upward force it experiences at a general point in the SHM cycle, when the displacement is x.

First the acceleration. Because we know the ball exhibits SHM we know that

\[ a =\ddot{x}= -\omega^2x \hspace{1cm}\mbox{\rm gives upward acceleration}\hspace{1cm} \]
You should satisfy yourself about the sign.

Now consider the force on the ball. The force has two contributions: the weight of the ball, mg, downwards; and the normal contact force from the bat,FB, upwards. Thus

\[ F = F_B-mg \hspace{1cm}\mbox{\rm gives net upward force}\hspace{1cm} \]
Newton’s 2nd Law then gives
-mω2x=F=FB-mg
which we rearrange in the form
\[ F_B = m(g-\omega^2x) = mg(1-\frac{x_m\omega^2}{g} \cos(\omega t)) \]

where we write the displacement as x=xmcos(ωt).

The figure shows how FB varies as a function of time for different values of the parameter

\[ \alpha= \frac{x_m\omega^2}{g} \]

The maxima and minima correspond (respectively) to points at which the bat-and-ball system is at the lowest and the highest points in the cycle. [You should think about this.]

Graphic - No title - batandball

When the bat and ball are on the point of separating the bat force will be zero. If the frequency of oscillation is fixed, and the amplitude gradually increased this condition will first be met at a point at the top of the cycle where FB has a minimum (and the cosine function is unity). The ’critical amplitude’ at which this occurs is then the solution of

\[ 1-\frac{x_m\omega^2}{g}=0 \hspace{0.5cm}\mbox{\rm giving}\hspace{0.5cm} x_m =\frac{g}{\omega^2} = 0.25\,m \]

Now suppose that the amplitude of oscillation is fixed and the frequency gradually increased. Separation (again at a point at the top of the cycle) will occur when

\[ \omega^2 x_m = g \hspace{0.5cm}\mbox{\rm implying}\hspace{0.5cm} f = \frac{1}{2\pi} \sqrt{\frac{g}{x_m}} = 2.2\, Hz \]