Problem — Rectangular torque and L(t)

Problem

A rigid disc is free to rotate about a horizontal axis. A torque $\tau(t)$ is applied to it, shaped as a positive rectangle of height $\tau_0 = 2{,}0\;\mathrm{N\,m}$ for $0 < t < 3{,}0\;\mathrm{s}$ and zero elsewhere. (a) Sketch $L(t)$ and $\omega(t)$ if $I = 0{,}50\;\mathrm{kg\,m^2}$ and the disc starts from rest. (b) What relationship is there between the area of the rectangle and the final angular momentum?

Solution

Angular momentum is the integral of torque: $\tau = \dfrac{dL}{dt}$, so $L(t)$ is the accumulated area under $\tau(t)$. With $\tau$ constant $= \tau_0$ for $3\;\mathrm{s}$, $L$ grows linearly up to $L_\text{fin} = \tau_0\cdot\Delta t = 2{,}0\cdot 3{,}0 = 6{,}0\;\mathrm{kg\,m^2/s}$ then stays constant (zero torque). The angular velocity is $\omega(t) = L(t)/I$, hence a ramp reaching $\omega_\text{fin} = 6{,}0/0{,}50 = 12\;\mathrm{rad/s}$.

On the left $L(t)$, on the right $\omega(t)$: both linear ramps while the torque is applied, then constant.

(b) The area of the rectangle $\tau(t)$ is exactly the angular impulse, equal to the change in angular momentum: $\ev{\text{area} = \tau_0\,\Delta t = \Delta L = L_\text{fin} = 6{,}0\;\mathrm{kg\,m^2/s}}$

Connections

Topics: Dinamica rotazionale Concepts: Momento angolare · Momento di una forza Skills: Lettura dei grafici Objects: Disco rotante