Question

The center of oscillation of a physical pendulum has this interesting property: If an impulse (assumed horizontal and in the plane of oscillation) acts at the center of oscillation, no oscillations are felt at the point of support. Baseball players (and players of many other sports) know that unless the ball hits the bat at this point (called the “sweet spot” by athletes), the oscillations due to the impact will sting their hands. To prove this property, let the stick in Fig. simulate a baseball bat. Suppose that a horizontal force $\overrightarrow{\mathbf{F}}$(due to impact with the ball) acts toward the right at P, the center of oscillation. The batter is assumed to hold the bat at O, the pivot point of the stick. (a) What acceleration does the point O undergo as a result of$\overrightarrow{\mathbf{F}}$? (b) What angular acceleration is produced by $\overrightarrow{\mathbf{F}}$about the center of mass of the stick? (c) As a result of the angular acceleration in (b), what linear acceleration does point O undergo? (d) Considering the magnitudes and directions of the accelerations in (a) and (c), convince yourself that P is indeed the “sweet spot.

Short answer

1. The acceleration of point O as a result of $\overrightarrow{F}$ is F/m.
2. The angular acceleration produced by $\overrightarrow{F}$about the center of mass of the stick is
   $\frac{2F}{mL}$ .
3. The linear acceleration of point O is zero.
4. Point P is indeed the ‘sweet spot’.

Step-by-step solution

The given data

• The figure of the stick simulating the baseball bat.
• P is the center of oscillation; C is center of mass and O is pivot point of the stick.
• The horizontal force $\overrightarrow{F}$ acts towards right at P.

Understanding the concept of kinematics

We can find the acceleration of point O as a result of force using Newton’s second law. The angular acceleration produced by force about the center of mass of the stick can be calculated from the torque acting on point P. From these two values, we can find the linear acceleration of point O using the relation between them. From parts a and c, we can interpret that point P is indeed the ‘sweet spot’.

Formulae:

The net force on a body using Newton’s second law, $F_{net}=ma$ (i)

Here, $F_{net}$is the net force acting on the body, is mass of the body, is acceleration.

The torque applied on a body,

role="math" localid="1660977854579" $$\begin{gathered} \tau =l\alpha \\ =r\times F \end{gathered}$$

Here, $\tau$ is torque, l is rotational inertia, $\alpha$is angular acceleration, r is the radius, F is force.

The radial acceleration of a body, $a=r\alpha$ (iii)

Here, a is linear acceleration, $\alpha$is angular acceleration, r is the radius,

$l=\frac{m{L}^2}{12}\left(\mathrm{iv}\right)$

The Moment of Inertia of the rod,

Here, l is rotational inertia, m is the mass of the body, L is the length of the rod.

(a) Calculation of acceleration of point O

Let’s assume that the batter exerts no force on the bat.

The net force acting on point O is $\overrightarrow{F}$Then the magnitude of its acceleration using equation (ii) is given as:

$a=\frac{F}{m}$

Therefore, the acceleration of point O as a result of $\overrightarrow{F}$is $\frac{F}{m}$.

(b) Calculation of angular acceleration

The torque acting on point P due $\overrightarrow{F}$using equation (ii) is given as:

$\tau =F\left(L_0-\frac{L}{2}\right)$

Since the distance between point O and P is $\frac{L}{2}$ but,$L_0=\frac{2}{3}L$

So,

$\begin{array}{l}\tau =F\left(\frac{2}{3}L-\frac{L}{2}\right)\\ =F\left(\frac{1}{6}L\right)\end{array}$

Now, using this above value and equation (iv) in equation (ii), we get the net angular acceleration as:

$$\begin{gathered} \alpha =\frac{F\left({\displaystyle \frac{1}{6}}L\right)}{{\displaystyle \frac{m{L}^2}{12}}} \\ =\frac{2F}{mL} \end{gathered}$$

Therefore, the angular acceleration produced by $\overrightarrow{F}$about the center of mass of the stick is $\frac{2F}{mL}$.

(c) Calculation of net acceleration of point O

The magnitude of linear acceleration of point O due to force $\overrightarrow{F}$is $\frac{F}{m}$.

Let’s assume right direction as positive.

Hence, the linear acceleration of point O due to linear acceleration using equation (iii) is given as:$a\text{'}=r\alpha$

Since the distance between Point C and O is L/2, the acceleration becomes

$a\text{'}=\frac{L}{2}\alpha$

The net linear acceleration of point O is given as:

$\begin{array}{l}{a}_O=\frac{F}{m}-\frac{L}{2}\alpha \\ =\frac{F}{m}-\frac{L}{2}\left(\frac{2F}{mL}\right)\\ =0\end{array}$x'

Therefore, the linear acceleration of point O is zero.

(d) Calculation of the sweet spot

From part a) and c) we can interpret that the net acceleration on point O is zero, it means that it is stationary when force is applied at point P. We assumed that the force exerted by the batter on the bat at point O is zero. Since point O is stationary, the force exerted on the batter’s hand is zero which helps to keep a good hold on the bat. This implies that the impulse at point P yields no oscillations at the point of support O and the oscillations do not sting the batter’s hand.

Therefore, point P is a sweet spot.