Impulse–Momentum Theorem in Rotational Form (College Board AP® Physics 1: Algebra-Based): Study Guide

Rate of change of angular momentum

• In terms of angular momentum, Newton's second law states:

The rate of change of angular momentum is equal to the net external torque exerted on an object (or system)

• This can be written as:

$\tau =\frac{∆L}{∆t}$

• Where:

  • $\tau$ = torque exerted on the system, measured in $\mathrm{N}·\mathrm{m}$

  • $∆L$ = change in angular momentum, measured in $\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}$

  • $∆t$ = time interval, measured in $\mathrm{s}$

• The magnitude of the change in angular momentum is the difference between an object or system's final and initial angular momenta:

change in angular momentum = final angular momentum − initial angular momentum

$∆L=L-L_0$

• The equation above can be used in situations where the rotational inertia of the body is not constant

• When the rotational inertia is constant:

${\tau }_{net}=\frac{∆L}{∆t}=\frac{L-L_0}{∆t}=\frac{I\omega -I{\omega }_0}{∆t}=\frac{I\left(\omega -{\omega }_0\right)}{∆t}=\frac{I∆\omega }{∆t}$

• Since angular acceleration $\alpha$ is equal to the rate of change of angular velocity, the equation becomes:

${\tau }_{net}=I\alpha$

• Where:

  • $I$ = rotational inertia of the body, measured in $\mathrm{kg}·{\mathrm{m}}^2$

  • $∆\omega$ = change in angular velocity, measured in $\mathrm{rad}/\mathrm{s}$

  • $\omega$ = final angular velocity, measured in $\mathrm{rad}/\mathrm{s}$

  • ${\omega }_0$ = initial angular velocity, measured in $\mathrm{rad}/\mathrm{s}$

  • $\alpha$ = angular acceleration, measured in $\mathrm{rad}/{\mathrm{s}}^2$

• More about this equation can also be found in the study guide for Newton’s second law in rotational form

Impulse–momentum theorem in rotational form

• In terms of angular momentum, Newton’s first law states:

The angular momentum of an object (or system) remains constant unless an external net torque acts upon it

• An angular impulse is exerted when an external torque is applied for a time

• Therefore, it must also change the angular momentum of the system

  • This is known as the rotational form of the impulse-momentum theorem

• The rotational form of the impulse-momentum theorem states that the angular impulse exerted on an object or rigid system is equal to the change in angular momentum

$\mathrm{angular}\mathrm{impulse}=∆L=\tau ∆t$

• Where:

  • angular impulse is measured in $\mathrm{N}·\mathrm{m}·\mathrm{s}$

  • $∆L$ = change in angular momentum, measured in $\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}$

  • $\tau$ = torque exerted on the system, measured in $\mathrm{N}·\mathrm{m}$

  • $∆t$ = time interval, measured in $\mathrm{s}$

• The equation will appear in this form on your equation sheet

• In calculations, a more useful form of this equation is

$∆L=\tau ∆t=I\left(\omega -{\omega }_0\right)$

• Where:

  • $I$ = rotational inertia of the body, measured in $\mathrm{kg}·{\mathrm{m}}^2$

  • $\omega$ = final angular velocity, measured in $\mathrm{rad}/\mathrm{s}$

  • ${\omega }_0$ = initial angular velocity, measured in $\mathrm{rad}/\mathrm{s}$

• This form can be derived from the equation for angular momentum $\left(L=I\omega \right)$

• The rotational form of Newton’s second law is a direct result of the rotational form of the impulse-momentum theorem applied to systems with constant rotational inertia

${\tau }_{net}=\frac{∆L}{∆t}=\frac{I∆\omega }{∆t}=I\alpha$

• Where:

  • $∆\omega$ = change in angular velocity, measured in $\mathrm{rad}/\mathrm{s}$

  • $\alpha$ = angular acceleration, measured in $\mathrm{rad}/{\mathrm{s}}^2$

• The rotational form of the impulse-momentum theorem tells us

  • for a given change in angular momentum, a small torque acting over a long time has the same effect as a large torque acting over a short time

  • for a constant torque, applying the torque over a longer time will lead to a greater change in angular momentum

  • for a specified time, a greater torque will lead to a greater change in angular momentum