Question

A softball player swings a bat, accelerating it from rest to 2.6 rev/srev/s in a time of 0.20 ss . Approximate the bat as a 0.90-kgkg uniform rod of length 0.95 mm, and compute the torque the player applies to one end of it.

Answer 1

Answer:

$\tau=22.13Nm$

Explanation:

information we have:

mass: $m=0.9kg$

lenght: $L=0.95m$

frequency: $f=2.6rev/s$

time: $t=0.2s$

and from the information we have we can calculate the angular velocity $\omega$. which is defined as

$\omega=2\pi f$

$\omega=2\pi (2.6rev/s)\\\omega=16.336 rev/s$

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Now, to calculate the torque

We use the formula

$\tau=I \alpha$

where $I$  is the moment of inertia and $\alpha$ is the angular acceleration

moment of inertia of a uniform rod about the end of it:

$I=\frac{1}{3}mL^2$

substituting known values:

$I=\frac{1}{3} (0.9kg)(0.95m)^2\\I=0.271kg/m^2$

for the torque we also need the acceleration $\alpha$ which is defined as:

$\alpha=\frac{\omega}{t}$

susbtituting known values:

$\alpha=\frac{16.336rev/s}{0.2s} \\\alpha=81.68rev/s^2$

and finally we substitute $I$ and  $\alpha$  into the torque equation $\tau=I \alpha$:$\tau=(0.271kg/m^2)(81.68rev(s^2)\\\tau=22.13Nm$