Question

If a body of mass 'm' is projected at an angle '$\theta$' with the horizontal with an initial velocity 'u', then the average torque on the body during the flight is (g - acceleration due to gravity)

• A. $\frac{mu^3 \sin \theta}{g}$
• B. $\frac{mu^3 \sin \theta}{2g}$
• C. $\frac{mu^2 \sin 2\theta}{g}$
• D. $\frac{mu^2 \sin 2\theta}{2}$

Hint:

The average torque can be found using the change in angular momentum over the time interval. For projectile motion from the origin, the initial angular momentum is zero. The average torque due to gravity acts only perpendicular to the plane of motion, and its magnitude is related to the initial velocity components.

Answer 1

The Correct Option is D

Step 1: Define average torque and angular momentum.
The average torque ($\vec{\tau}_{avg}$) acting on a body is equal to the total change in angular momentum ($\Delta \vec{L}$) divided by the total time taken ($\Delta t$).
\[ \vec{\tau}_{avg} = \frac{\Delta \vec{L}}{\Delta t} = \frac{\vec{L}_f - \vec{L}_i}{T} \] Here, $T$ is the total time of flight. Angular momentum ($\vec{L}$) is defined as $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v}$, where $\vec{r}$ is the position vector from the origin, $m$ is the mass, and $\vec{v}$ is the velocity. We choose the origin as the point of projection. Step 2: Calculate initial angular momentum ($\vec{L}_i$).
At the point of projection (initial time $t=0$), the position vector $\vec{r}_i = \vec{0}$.
Therefore, the initial angular momentum $\vec{L}_i = \vec{r}_i \times m\vec{u} = \vec{0} \times m\vec{u} = \vec{0}$. Step 3: Calculate the time of flight ($T$).
For a projectile launched with initial velocity $u$ at an angle $\theta$ with the horizontal, the time of flight is given by: \[ T = \frac{2u \sin \theta}{g} \] Step 4: Calculate final position and velocity at time of flight ($T$).
At time $T$, the projectile returns to the horizontal level (ground, $y=0$).
The horizontal distance traveled (range) is:
\[ x_f = (u \cos \theta) T = (u \cos \theta) \left(\frac{2u \sin \theta}{g}\right) = \frac{2u^2 \sin \theta \cos \theta}{g} = \frac{u^2 \sin (2\theta)}{g} \] So, the final position vector is $\vec{r}_f = \frac{u^2 \sin (2\theta)}{g} \hat{i}$. (assuming projection is from origin, and horizontal is x-axis) The final velocity components are:
$v_{fx} = u \cos \theta$
$v_{fy} = u \sin \theta - gT = u \sin \theta - g\left(\frac{2u \sin \theta}{g}\right) = u \sin \theta - 2u \sin \theta = -u \sin \theta$ So, the final velocity vector is $\vec{v}_f = (u \cos \theta)\hat{i} - (u \sin \theta)\hat{j}$. Step 5: Calculate final angular momentum ($\vec{L}_f$).
$\vec{L}_f = \vec{r}_f \times m\vec{v}_f$ $\vec{L}_f = \left( \frac{u^2 \sin (2\theta)}{g} \hat{i} \right) \times m \left( (u \cos \theta)\hat{i} - (u \sin \theta)\hat{j} \right)$ Using the cross product properties ($\hat{i} \times \hat{i} = \vec{0}$ and $\hat{i} \times \hat{j} = \hat{k}$): $\vec{L}_f = m \frac{u^2 \sin (2\theta)}{g} (u \cos \theta)(\hat{i} \times \hat{i}) - m \frac{u^2 \sin (2\theta)}{g} (u \sin \theta)(\hat{i} \times \hat{j})$ $\vec{L}_f = \vec{0} - m \frac{u^3 \sin (2\theta) \sin \theta}{g} \hat{k}$ $\vec{L}_f = - \frac{m u^3 \sin (2\theta) \sin \theta}{g} \hat{k}$ Step 6: Calculate the average torque.
\[ \vec{\tau}_{avg} = \frac{\vec{L}_f - \vec{L}_i}{T} = \frac{- \frac{m u^3 \sin (2\theta) \sin \theta}{g} \hat{k} - \vec{0}}{\frac{2u \sin \theta}{g}} \] \[ \vec{\tau}_{avg} = \frac{- \frac{m u^3 \sin (2\theta) \sin \theta}{g}}{\frac{2u \sin \theta}{g}} \hat{k} \] Cancel common terms ($\frac{u \sin \theta}{g}$): \[ \vec{\tau}_{avg} = - \frac{m u^2 \sin (2\theta)}{2} \hat{k} \] The magnitude of the average torque is $\frac{m u^2 \sin (2\theta)}{2}$. The final answer is $\boxed{\frac{mu^2 \sin 2\theta}{2}}$.