Question

A circle is spinning at 220 revolution per minute. It stops spinning in 5.25 revolutions. What is the angular acceleration?

Answer 1

idk sry

Answer 2

Hello. Do you remember your kinematic equations? You may have also been taught in studying rigid body rotations that all the equations you learned in linear kinematics and dynamics have exact analogues in the rotational domain. In other words, if you replace the linear variables with their rotational counterpart variables, you can arrive at the rotational formulas.

If you replace:
d (linear displacement) with theta (angular displacement),
v (linear velocity) with omega (angular velocity),
a (linear acceleration) with alpha (angular acceleration)
m (mass) with I (moment of inertia)
p (momentum) with L (angular momentum)
F (force) with tau (torque)
you will arrive at the rotational equation you need.

You may recall one of the kinematic equations:
\[v _{f}^{2} = v _{i}^{2} + 2 a d\]
If I make the substitutions that I suggested above, I get
\[\omega _{f}^{2} = \omega _{i}^{2} + 2 \alpha \theta\]
where omega_f and omega_i are final and initial angular velocities respectively; alpha is the angular acceleration; and theta is the angular displacement. In your case,
\[\omega _{f} = 0\]
\[\omega _{i} = \frac{ 2\pi(220 rev/\min) }{ 60 \min/s}\]\[\theta = (5.25 rev)(2\pi)\]
Solve for alpha. You should note that the angular acceleration should be negative.