a cheerleader has a baton that is 1 meter long and has a 100 gram mass on both ends another cheerleader has a baton that is 0.5 meterlong and has 100 gram masson both ends. Neglecting the mass of the rod holding the masses together, which baton will be easier to spin when grasped at the center?
To determine which baton will be easier to spin when grasped at the center, we need to consider the concept of rotational inertia or moment of inertia, which determines how resistant an object is to changes in its rotation.
The formula for rotational inertia is given by:
I = m * r^2
where I is the moment of inertia, m is the mass, and r is the distance from the axis of rotation.
In this case, we have a baton with masses on both ends. Since the baton is grasped at the center, the axis of rotation will also be at the center.
For the first baton with a length of 1 meter:
The distance from the axis of rotation to each mass on either end is 0.5 meters (half the length of the baton). Therefore, r = 0.5 m.
Using the moment of inertia formula, the moment of inertia for the first baton (I1) can be calculated as:
I1 = (m * r^2) + (m * r^2)
= m * (r^2 + r^2)
= m * (2r^2)
= m * 2 * (0.5^2)
= m * 2 * (0.25)
= 0.5 * m
For the second baton with a length of 0.5 meters:
In this case, the distance from the axis of rotation to each mass on either end is 0.25 meters (half the length of the baton). So, r = 0.25 m.
Using the moment of inertia formula, the moment of inertia for the second baton (I2) can be calculated as:
I2 = (m * r^2) + (m * r^2)
= m * (r^2 + r^2)
= m * (2r^2)
= m * 2 * (0.25^2)
= m * 2 * (0.0625)
= 0.125 * m
Comparing the two moment of inertia values:
I1 = 0.5 * m
I2 = 0.125 * m
Since I2 is smaller than I1, the moment of inertia (resistance to rotation) for the second baton is lower. Therefore, the baton with a length of 0.5 meters will be easier to spin when grasped at the center compared to the baton with a length of 1 meter.