A softball player swings a bat, accelerating it from rest to 8 rev/s in a time of 0.3 s. Find the torque the player applies to one end of the bat. Approximate the bat as a 2.2 kg uniform rod of length 0.95 m.
alpha = ( 8 * 2 pi radians/s )/.3s
= 168 radians/s^2
torque = I alpha
I = (1/3) m L^2 = (1/3)(2.2)(.95)^2 = .662 kg m^2
so
.662 * 168 kg m^2/s^2 = 111 N m
To determine the torque applied to the bat, we need to calculate the moment of inertia of the bat and use the equation for torque.
The moment of inertia of a uniform rod about its center of mass can be calculated using the formula:
\(I = \frac{1}{12} m L^2\)
Where:
- \(I\) is the moment of inertia
- \(m\) is the mass of the rod
- \(L\) is the length of the rod
Given the mass of the bat (\(m = 2.2 \, \text{kg}\)) and the length of the bat (\(L = 0.95 \, \text{m}\)), we can substitute these values into the formula to find the moment of inertia (\(I\)).
\(I = \frac{1}{12} \times 2.2 \, \text{kg} \times (0.95 \, \text{m})^2\)
Now that we have the moment of inertia, we can use the equation for torque:
\(T = I \alpha\)
Where:
- \(T\) is the torque
- \(I\) is the moment of inertia
- \(\alpha\) is the angular acceleration
Given the final angular velocity (\(8 \, \text{rev/s}\)) and the time taken to reach that velocity (\(0.3 \, \text{s}\)), we can calculate the angular acceleration (\(\alpha\)) using the formula:
\(\alpha = \frac{\Delta \omega}{\Delta t}\)
Where:
- \(\Delta \omega\) is the change in angular velocity
- \(\Delta t\) is the change in time
Since the bat starts from rest (\(\omega_i = 0\)), the change in angular velocity is equal to the final angular velocity (\(\Delta \omega = 8 \, \text{rev/s}\)).
Now, we can substitute the values of the moment of inertia (\(I\)) and the angular acceleration (\(\alpha\)) into the torque equation (\(T = I \alpha\)) to find the torque (\(T\)).