I have to calculate the arc length in radians that a circle spinning at speed will travel when it decelerates to $0$.
I have the initial angular velocity in rad/s, the radius in meters, the mass in kg, and the decelerating Force acting on it in N.
I know that in a linear problem it would be initial velocity squared over 2 x acceleration but not sure what it would be here as we're dealing with torque and angular components. Can someone walk me through the equations to use?
I think the equation I should be using is $S = 0.5 a t^2$ where t is found with angular velocity over deceleration, but I don't know how to get deceleration using the torque equations instead of the linear $F=ma$.
Best Answer
So indeed, the torque $\vec \tau = \vec r \times \vec F$ slows this thing down, and assuming that the force is never radial but only tangential, this is just $\tau = r ~ F$ where r is the radius of the circle and $F$ is the magnitude of the braking force.
Newton's law $\frac{d\vec p}{dt} = \sum_i \vec F_i$ becomes for angular problems $\frac{d\vec L}{dt} = \sum_i \vec \tau_i$, where $\vec L$ is the angular momentum. Just like how $\vec p = m \vec v$ you generally have a "moment of inertia" I such that $|L| = I \omega$. (Unlike mass, for complicated rotations the angular momentum vector may not point along the same line as the rotation vector, leading to a "moment of inertia tensor.")
In your case you should determine whether the circle is "filled in uniformly" or whether all of its mass is concentrated at its edge or what; that will give you a formula for its moment of inertia $I$ about its center. For example if the mass is concentrated at the edge, then $I = mr^2$.
Since $I$ is constant with respect to time, we just have $\frac{dL}{dt} = I \frac{d\omega}{dt}$ and hence we have an exact copy of the free-fall equations: for a constant torque $\tau$, $$\theta(t) = \theta_0 + \omega_0 t + \frac{1}{2}~\frac{\tau}{I} ~t^2.$$This parameter $\tau/I$ is sometimes called the "angular acceleration."