Question

A particle of mass 1kg, initially at rest, starts sliding down from the top of a frictionless inclined plane of angle \(\frac{𝜋}{6}\)\(\frac{\pi}{6}\) (as schematically shown in the figure). The magnitude of the torque on the particle about the point O after a time 2seconds is ______N-m. (Rounded off to nearest integer) 

(Take g = 10m/s²)

Answer 1

Correct Answer: 85

Given Data 

• Mass of the particle, m = 1 kg
• Angle of the inclined plane, θ = π/6
• Initial velocity, u = 0 m/s
• Time, t = 2 seconds
• Gravitational acceleration, g = 9.8 m/s²

Step 1: Calculate the Acceleration Along the Inclined Plane

The component of gravitational acceleration along the incline is:

a = g sinθ = 9.8 sin(π/6)

Substituting sin(π/6) = 0.5:

a = 9.8 × 0.5 = 4.9 m/s²

Step 2: Calculate the Distance Traveled by the Particle

Using the equation of motion:

s = ut + (1/2) a t²

Since u = 0, substitute a = 4.9 m/s² and t = 2 seconds:

s = 0 + (1/2) × 4.9 × (2)²

s = (1/2) × 4.9 × 4 = 9.8 m

Step 3: Calculate the Torque About Point O

The perpendicular distance from point O to the line of action of the gravitational force is:

r⊥ = s cosθ

Substituting s = 9.8 m and cos(π/6) = √3/2 ≈ 0.866:

r⊥ = 9.8 × (√3/2) ≈ 9.8 × 0.866 = 8.48 m

Step 4: Calculate the Gravitational Force

The gravitational force acting on the particle is:

F = mg = 1 × 9.8 = 9.8 N

Step 5: Calculate the Torque

The torque about point O is:

τ = r⊥ × F

Substituting r⊥ = 8.48 m and F = 9.8 N:

τ = 8.48 × 9.8 ≈ 83.1 N·m

Final Answer

The torque about point O is 83.1 N·m.