A string is wrapped around the rim of a wheel of moment of inertia 0.40 kgm² and radius 10 cm. The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of 40 N. The angular velocity of the wheel after 10 s is x rad / s, where x is __________.

Solution

Physics sir · 3 steps

The pull on the rim gives a constant torque, so get α from τ = Iα and use ω = αt.

1. Torque from the string. The string leaves the rim along a tangent, so the force is perpendicular to the radius. With r = 10 cm = 0.1 m, the torque is τ = Fr = 40 × 0.1 = 4 N m.

   α = τ/I = 4/0.40 = 10 rad s⁻²

2. Constant angular acceleration. The force and the radius do not change, so α stays constant. The wheel starts from rest, so ω = ω₀ + α t = 0 + 10 × 10.
3. So. ω = 100 rad s⁻¹, so x = 100.

The common slip: 10000. Using the radius as 10 instead of 0.1 m makes the torque 400 N m and gives 10 000, not 100. Convert cm to m first.

The idea to keep. A string pulled off a rim gives torque Fr. Rotation then works just like linear motion: τ = Iα plays the role of F = ma, and ω = ω₀ + α t plays the role of v = u + at.

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