Q.To maintain a rotor at a uniform angular speed of 120 rad/s, engine needs to transmit a torque of 180 Nm. What is the power required by the engine?

Power Against Gravity: The Intuition

Imagine you're lifting a bucket of water from a well. The bucket is heavy — gravity is pulling it down with a force equal to its weight. To lift it, you must apply an upward force that exactly cancels gravity. If you lift it slowly, you feel the strain for a long time. If you yank it up quickly, you feel a burst of effort, but it's over fast.

That "burst of effort per unit time" is what we call power. When you're working against a constant force like gravity, power tells you how fast you're doing that work.

The Precise Statement

When an object moves vertically against gravity (upward), the force you must supply is at least equal to the object's weight:

where $m$ is mass and $g$ is acceleration due to gravity ($9.8\ \text{m/s}^2$ on Earth).

The work done to lift it through a height $h$ is:

Now, power $P$ is work per unit time. If you lift it in time $t$:

But $\frac{h}{t}$ is just the upward speed $v$ (assuming constant speed). So we get the compact form:

This is power against gravity — the rate at which you must supply energy to lift a mass $m$ at constant speed $v$ against the pull of gravity.

What This Really Means

• It's a minimum. If you accelerate the object upward, you need even more force (Newton's second law), and hence more power. The formula $P = mgv$ assumes you're lifting at steady speed — no acceleration.
• Direction matters. If the object moves downward at constant speed, gravity does the work, and you (or a brake) must absorb power. The formula still gives the magnitude, but the sign flips.
• It's independent of path. Only the vertical speed matters. Whether you lift straight up or along a ramp, the power against gravity depends only on the vertical component of velocity.

Common Mistake to Avoid …