A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by force of 10 N as shown in figure. The established torque produces an angular acceleration of 2 rad / s². Moment of intertia of the wheel is___________ kg m². (Acceleration due to gravity =10 m / s² )

Solution

Physics sir · 3 steps

The string pulls along a tangent, so the lever arm is the full radius. Then use torque equals I alpha.

1. Find the lever arm. The string leaves the rim along a tangent, so the 10 N force is perpendicular to the radius at that point. Its perpendicular distance from the axle is the radius, r = 0.2 m.
2. Torque about centre. τ = F r = 10 × 0.2 = 2 N m. The axle force passes through the centre, so it gives no torque.

   τ = Iα ⇒ I = τ/α = (2 N m)/(2 rad s⁻²)

3. So. I = 1 kg m², so the answer is 1. The value of g is not needed, because the pull is already given as a force in newtons.

The common slip: 10. Treating 10 N as a 10 kg hanging mass and multiplying by g makes the torque 20 N m and gives I = 10.

The idea to keep. For a string wrapped on a rim, torque is always tension times radius. Then τ = Iα about the fixed axle gives any one of the three quantities.

Answer from the NTA final answer key. Solution by PrepWiser.