Question

The range of a projectile of weight \( W \) is \( R \). The average torque on the projectile between the initial and final positions \( P \) and \( Q \) about the point of projection is:

• A. \( \frac{WR}{2} \)
• B. \( \frac{WR}{3} \)
• C. \( \frac{WR}{4} \)
• D. \( WR \)

Hint:

The average torque is half the product of the weight and range of the projectile.

Answer 1

The Correct Option is A

The torque on the projectile is defined as the moment of the force at a given point. The average torque between two points \( P \) and \( Q \) on the projectile's path is given by: \[ \text{Average Torque} = \frac{1}{2} \times \text{Force} \times \text{Distance} \] Here, the force on the projectile is its weight \( W \) and the distance between \( P \) and \( Q \) is the range \( R \). Therefore, the average torque is: \[ \text{Average Torque} = \frac{WR}{2} \] Thus, the average torque between the initial and final positions \( P \) and \( Q \) is \( \frac{WR}{2} \).

Answer 2

Step 1: Understand the physical situation.
A projectile is launched and follows a parabolic path from point \( P \) (start) to point \( Q \) (end). The range \( R \) is the horizontal distance between these two points. The weight \( W \) acts vertically downward through the center of mass throughout the motion.

Step 2: Torque due to weight about the point of projection.
Torque \( \tau \) is given by: \[ \tau = r \times F = r F \sin\theta \] In this case:
- \( r \) is the position vector from the origin (point of projection) to the projectile’s instantaneous position.
- \( F = W \) (weight acting downward)
- The torque due to weight acts about the horizontal displacement from the origin.

Step 3: Average torque over the flight.
Since the vertical force (weight) remains constant and the horizontal displacement ranges from 0 to \( R \), the average horizontal lever arm is \( \frac{R}{2} \).

So, average torque: \[ \tau_{\text{avg}} = W \times \frac{R}{2} \]
Final Answer: \( \boxed{\frac{WR}{2}} \)