Angular Impulse

From linear impulse to angular impulse

• In linear motion, the resultant force on a body is the rate of change of linear momentum: $$F = \frac{\Delta p}{\Delta t}$$
• This leads to the definition of linear impulse: $\Delta p = F\Delta t = \Delta(mv)$.
• Similarly, the resultant torque on a body is the rate of change of angular momentum: $$\tau = \frac{\Delta L}{\Delta t}$$
• This leads to the definition of angular impulse: $$\Delta L = \tau\Delta t = \Delta(I\omega)$$

Key features of angular impulse

• This equation requires the use of a constant resultant torque. If the torque changes, an average must be used.
• Fundamentally, angular impulse describes the effect of a torque acting over a time interval. This means a small torque acting over a long time has the same effect as a large torque acting over a short time.

Angular impulse on a torque-time graph

• The area under a torque-time graph is equal to the angular impulse, or the change in angular momentum.
• This is because: area = base $\times$ height = torque $\times$ time = $\tau\Delta t$.
• When the torque is not constant, you must calculate the area geometrically (using triangles, rectangles, or counting squares).

Worked example: torque-time graph

• An object with $I = 6.0$ kg m$^2$ rotates clockwise at 2.0 rad s$^{-1}$ at $t = 0$. A torque-time graph shows a positive triangle (area = 15 N m s) followed by a negative rectangle (area = $-10$ N m s).
• Total change in angular momentum: $\Delta L = 15 - 10 = 5$ N m s.
• Using $\Delta L = I(\omega_f - \omega_i)$ with $\omega_i = -2.0$ rad s$^{-1}$ (clockwise is negative): $$6(\omega_f - (-2)) = 5$$ $$6\omega_f + 12 = 5$$ $$\omega_f = \frac{-7}{6} = -1.17 \text{ rad s}^{-1}$$
• The negative sign means the object is still rotating clockwise at 1.17 rad s$^{-1}$.