Rotational Energy, Work and Power

Use forces and torques when you need acceleration or contact forces. Use energy when you need a speed after a displacement and can account for the work.

Why translational and rotational energy add

For particles making up a rigid body, write $𝐯_{i} = 𝐯_{C} + 𝛚 \times 𝛒_{i}$ , where $𝛒_{i}$ is measured from the centre of mass. Expand $\frac{12}{\sum} m_{i} v_{i}^{2}$ . The cross term vanishes because $\sum m_{i} 𝛒_{i} = 0$ . For planar motion this leaves

K = \frac{12}{M} v_{C}^{2} + {\frac{12}{I}}_{C} ω^{2} .

This formula describes one body in general planar motion. If it instead rotates about a fixed axis through $O$ , substitute $v_{C} = ω d$ and $I_{O} = I_{C} + M d^{2}$ to obtain $K = {\frac{12}{I}}_{O} ω^{2}$ . These are two equivalent descriptions, not two energies to add together.

A force acting on a body rotating about a fixed axis does infinitesimal work $d W = τ_{z} d θ$ . Summing torques and integrating gives

W_{n e t} = \int_{θ_{i}}^{θ_{f}} τ_{z} (θ) d θ = {\frac{12}{I}}_{z} ω_{f}^{2} - {\frac{12}{I}}_{z} ω_{i}^{2}, P = τ_{z} ω .

A torque–angle graph therefore has work as its signed area. The rectangle $τ Δ θ$ works only for constant torque. Torque and energy share the dimensions $N m$ , but torque is not “energy”: one describes rotational influence, the other a scalar transfer or state quantity. Report torque in $N m$ and work in joules.

For an individual external force on a rigid body, the useful power decomposition is $𝐅 \cdot 𝐯_{C} + 𝛕_{C} \cdot 𝛚$ . A force can decrease translational energy and increase rotational energy simultaneously. Always ask for the velocity of its actual point of application before declaring that the force does or does not do work.

Checkpoint: two wheels receive the same torque through the same angle. Which gains more energy?

They gain the same energy, $Δ K = τ Δ θ$ , even if their moments of inertia differ. Starting from rest, the smaller-inertia wheel reaches the larger angular speed because $ω = \sqrt{2 τ Δ θ / I}$ . But if the same constant torque instead acts for the same time, their angular-momentum changes are equal and their energy gains are $τ^{2} t^{2} / (2 I)$ . Read carefully whether a problem specifies equal time, angle, torque, or angular impulse: these are different experiments.

The wheel keeps accelerating after its torque starts decreasing

A wheel has fixed-axis inertia $2.0 k g m^{2}$ and initial angular velocity $+ 2.0 r a d s^{- 1}$ at $θ = 0$ . Its net torque is $τ_{z} = (8 - 2 θ) N m$ , with $θ$ expressed in radians. Find its speed at $θ = 3$ , the angle of maximum speed before the first stop, and the first stopping angle.

Hint: obtain kinetic energy as a function of angle

A decreasing positive torque still increases kinetic energy. Speed becomes maximal when the torque changes from positive to negative along the forward motion.

Full solution

K (θ) = \frac{12}{(} 2) (2^{2}) + \int_{0}^{θ} (8 - 2 ϑ) d ϑ = 4 + 8 θ - θ^{2} J .

At $θ = 3$ , $K = 19 J$ ; since $K = ω^{2}$ numerically in the stated SI units, $ω = + \sqrt{19} = 4.36 r a d s^{- 1}$ . Maximum energy occurs at $d K / d θ = 8 - 2 θ = 0$ , giving $θ = 4 r a d$ and $ω_{\max} = \sqrt{20} = 4.47 r a d s^{- 1}$ . The first stop satisfies

4 + 8 θ - θ^{2} = 0, θ_{s t o p} = 4 + \sqrt{20} = 8.47 r a d .

The negative root is outside the initial forward trajectory. After the stop, the negative torque would reverse the wheel; the “first stop” is not a permanent rest state.

Check: At $θ = 0$ the expression returns $K = 4 J$ . At $θ = 4$ , torque and acceleration are zero while speed is greatest.

The falling rod at the bottom

The rod from Example 10 swings from horizontal to vertically downward. Find its angular speed and hinge reaction at that instant. The hinge is frictionless, and air resistance is negligible.

Hint: the hinge does no work, but its force need not be small

At the bottom, the centre has upward centripetal acceleration. Gravity has zero hinge torque there.

Full solution

The centre falls by $L / 2 = 1 m$ . Since the hinge point is stationary, its reaction does zero work:

M g \frac{L}{2} = \frac{12}{(\frac{M L^{2}}{3})} ω^{2}, ω = \sqrt{\frac{3 g}{L}} = \sqrt{15} = 3.87 r a d s^{- 1} .

The bottom position has zero angular acceleration because gravity's line of action passes through the hinge. Nevertheless, the centre accelerates towards the hinge:

a_{C} = ω^{2} \frac{L}{2} = 15 m s^{- 2}, R - M g = M a_{C} .

Thus $R = 30 + 45 = 75 N$ , upward. Its horizontal component is zero at this instant.

Check: Generally $R = 5 M g / 2$ . Compare with the initial $M g / 4$ : “zero angular acceleration” does not imply zero acceleration of the centre or a reaction equal to weight.

A useful refusal: Energy conservation alone cannot determine every contact force. An ideal hinge can supply a large reaction while doing no work; find such forces from acceleration and Newton's law.

In 60 seconds

Why translational and rotational energy add

The wheel keeps accelerating after its torque starts decreasing

The falling rod at the bottom