Question

Find the angular impulse applied to the particle in the 4 sec.

• A. $10{\text{kg-m}}^2/\text{sec}$
• B. $40{\text{kg-m}}^2/\text{sec}$
• C. $20{\text{kg-m}}^2/\text{sec}$
• D. $5{\text{kg-m}}^2/\text{sec}$

Answer 1

The correct option is C $20{\text{kg-m}}^2/\text{sec}$

From the fundamental equation:
$\tau =\frac{dL}{dt}$
$\therefore \Delta L={\int }_{L_i}^{L_f}dL=\int \tau dt$
$\Delta L=\text{Change in angular momentum}$
$\int \tau dt=\text{Angular impulse}$
$\Rightarrow$Angular impulse applied to the particle in the time interval of $4\text{s}$ is given by area under torque-time i.e $(\tau -t)$ graph.
Angular impulse $\left(J\right)$ = Area of triangle $ABC$
$=\frac{1}{2}\times 4\times 10$
$=20{\text{kg-m}}^2/\text{sec}$