Question

A baseball bat could be considered a uniform rod. It ha a mass of 0.97 kg and a length of 97 cm. If a player accelerates it from rest to 2.8 rev/s in 0.20s, how much torque is applied?

Answer 1

Answer:

Applied torque, T = 26.74 N-m

Given:

Mass of rod, M = 0.97 kg

Length of the rod, L = 97 cm = 0.97 m

Change in angular velocity, $\Delta \omega = 2.8 rev/s = 2.8\times 2\pi = 17.59 rad/s$

time, t = 0.20 s

Solution:

Now, we know that torque is given by:

$T = I\alpha$                          (1)

where

I = moment of inertia of the rod

$\alpha$  = angular acceleration

Using perpendicular axis formula to calculate the moment of inertia:

$I = \frac{ML^{2}}{12}$                    

Now, the base ball bat will be held from one end, therefore,  the moment of inertia about the axis at the end is given by:

$I = \frac{ML^{2}}{12} + \frac{ML^{2}}{4} = \frac{ML^{2}}{3}$

$I = \frac{0.97\times (0.97)^{2}}{3} = 0.304 kg-m^{2}$

Now, angular acceleration can be calculated as:

$\alpha = \frac{\Delta \omega}{t} = \frac{17.59}{0.20} = 87.95 rad/s^{2}$

Now, calculation of torque using eqn (1):

$T = 0.304\times 87.95 = 26.74 N-m$

Answer 2

Answer:

The torque is 26.75 N-m.

Explanation:

Given that,

Mass = 0.97 kg

Length = 97 cm

Time = 0.20 s

Angular speed $\omega= 2.8 rev/s = 2.8\times2\pi\ rad/s$

We need to calculate the moment of inertia of rod at one end

Using formula of the moment of inertia

$I=\dfrac{Ml^2}{3}$

Put the value into the formula

$I=\dfrac{0.97\times(0.97\times10^{-2})^2}{3}$

$I=0.3042\ kg m^2$

We need to calculate the angular acceleration

Using formula of angular acceleration

$\alpha=\dfrac{\Delta \omega}{\Delta t}$

Put the value into the formula

$\alpha=\dfrac{2.8\times2\pi}{0.20}$

$\alpha=87.964\ rad/s^2$

We need to calculate the torque

Using formula of torque

$\tau=I\times\alpha$

Put the value into the formula

$\tau=0.3042\times87.964$

$\tau=26.75\ N-m$

Hence, The torque is 26.75 N-m.