91影视

• 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 $\stackrel{鈫�}{F}$ acts towards right at P.

We can find the acceleration of point O as a result of force using Newton鈥檚 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 鈥榮weet spot鈥�.

Formulae:

The net force on a body using Newton鈥檚 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" $$\begin{gathered} 蟿=l伪 \\ =r脳F \end{gathered}$$

Here, $蟿$ is torque, l is rotational inertia, $伪$is angular acceleration, r is the radius, F is force.

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

Here, a is linear acceleration, $伪$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.

Let鈥檚 assume that the batter exerts no force on the bat.

The net force acting on point O is $\stackrel{鈫�}{F}$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 $\stackrel{鈫�}{F}$is $\frac{F}{m}$.

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

$蟿=F\left(L_0-\frac{L}{2}\right)$

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

So,

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

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

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

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

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

Let鈥檚 assume right direction as positive.

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

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

$a\text{'}=\frac{L}{2}伪$

The net linear acceleration of point O is given as:

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

Therefore, the linear acceleration of point O is zero.

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鈥檚 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鈥檚 hand.

Therefore, point P is a sweet spot.