Mechanical power developed by a rotating shaft: If a shaft transmits torque T (in N·m) at a speed of N r.p.m., what is the general expression for the mechanical power developed by the torque?

Introduction / Context:
Converting rotational motion to power is central in machine design. The standard power–torque–speed relation allows engineers to size motors, couplings, and shafts and to convert between units (W, kW, hp) reliably.

Given Data / Assumptions:

• Torque T in newton–metres (N·m).
• Speed N in revolutions per minute (r.p.m.).
• Angular speed ω relates to N by ω = 2πN/60 rad/s.

Concept / Approach:

Mechanical power for rotation is P = torque * angular speed = T * ω. Substituting ω in terms of N gives the widely used formula with the factor 2π/60 when N is in r.p.m. If N were in revolutions per second, P = 2π N T would apply directly.

Step-by-Step Solution:

Verification / Alternative check:

Unit check: N·m * rad/s = joules/second = watts. For kW, divide by 1000; for horsepower (metric), hp ≈ P/735.5.

Why Other Options Are Wrong:

(b) and (c) are dimensionally inconsistent for N in r.p.m.; (d) cannot be true since the alternatives disagree; (e) is correct only if N is in revolutions per second, not r.p.m.

Common Pitfalls:

Forgetting the 60 factor when using r.p.m.; mixing N in r.p.s. with N in r.p.m.; confusing torque units (N·m vs kgf·m).

Final Answer:

P = (2π N T) / 60 watts