Power in Rotational Motion – Torque–Speed Relation A body is acted upon by a constant torque T (N·m) and rotates with angular speed ω (rad/s). What is the power developed?

Introduction / Context:
Converting between torque, angular speed, and power is essential in rotating machinery (motors, gearboxes, turbines). Knowing the exact relation avoids unit mistakes when sizing drives and couplings.

Given Data / Assumptions:

• Torque T in newton·metre (N·m).
• Angular speed ω in rad/s.
• Steady rotation (instantaneous power relation).

Concept / Approach:
Mechanical power is the rate of doing work. In rotation, the elemental work is dW = T dθ. Dividing by time gives power P = dW/dt = T (dθ/dt) = T ω. This yields power in watts when SI units are used.

Step-by-Step Solution:
Start: dW = T dθ. Differentiate: P = dW/dt = T * dθ/dt. Recognize dθ/dt = ω (rad/s). Therefore, P = T * ω (watts).

Verification / Alternative check:
Dimensional analysis: N·m × 1/s = N·m/s = watt, confirming SI consistency. When speed is given in revolutions per minute n (rpm), use ω = 2π n / 60 to obtain P = T * 2π n / 60 (watts) or P(kW) = (2π T n) / 60,000.

Why Other Options Are Wrong:
T * ω / 60: would be correct only if ω were in rpm; here ω is in rad/s already. T * ω / 75 or / 4500 in kW: arbitrary constants; not general. 2π * T * ω: double counts 2π since ω already includes radians.

Common Pitfalls:
Mixing rpm and rad/s; always convert rpm to rad/s before using P = T ω. Confusing torque units (N·m) with energy units (J); although numerically identical units appear, their physical meanings differ.

Final Answer:
T * ω (in watts)