Force Twirling a Baton

## Question:

A baton consists of a thin uniform rod (length=0.81m, mass=0.13kg), to either end which is attached to a very small ball (mass=0.13kg). A baton twirler can bring this baton from rest up to an angular speed of 2.0 rev/s in 0.40s by grasping the baton at its center and flicking her wrist. This causes a force to be applied to the baton at a point 2.0 cm away from the rotational axis through the center of the rod. Find the magnitude of the force.
## Answer:

Let's work backward from what is needed to see how what we know may turn out to be useful.
You need the force. The force is the torque divided by the **moment arm**. The torque is the moment of inertia times the angular acceleration. The angular acceleration is the rate of change in the angular velocity. The rate of change in angular velocity is the **change in velocity** divided by **the time to reach the final velocity**. The moment of inertia of the whole baton is the sum of the moment of inertia of its parts. The moment of inertia of a thin rod rotated about its center is its **mass** times the square of its **length** divided by 12. The moment of inertia of a small sphere about a point is its **mass** times its **distance from the point** squared.

You may notice that as we write out this string of facts we come across the stuff we have been given. I used bold face in the preceding paragraph to indicate things I think we know.

The torque moment arm is given as 0.02 meters. The change in angular velocity is 2 revolutions per second which is 4pi radians per second. The time to spin up is given as 0.40 seconds. The mass of the rod is given as 0.13 kg and its length is 0.81 meters. The mass of each ball is 0.13 kg and each lies a distance of 0.405 meters from the axis of rotation.

I think if you plug the numbers into the recipe we have in the first paragraph you will find the force.