Problem — Three regimes of ω(t)

Problem

The graph shows $\omega(t)$ of a disc starting from rest, accelerating for $4\;\mathrm{s}$ up to $20\;\mathrm{rad/s}$, holding that speed for $3\;\mathrm{s}$, then slowing to zero in $5\;\mathrm{s}$. Calculate $\alpha$ in each phase and describe the net torque.

Solution

The angular acceleration is the slope of $\omega(t)$; the net torque is $\tau = I\alpha$, with the same sign as $\alpha$.

• Phase 1 ($0$–$4\;\mathrm{s}$): $\alpha = \dfrac{20-0}{4} = 5\;\mathrm{rad/s^2}$ → positive torque (accelerating).
• Phase 2 ($4$–$7\;\mathrm{s}$): $\omega$ constant, $\alpha = 0$ → zero net torque.
• Phase 3 ($7$–$12\;\mathrm{s}$): $\alpha = \dfrac{0-20}{5} = -4\;\mathrm{rad/s^2}$ → braking torque. $\ev{\alpha_1 = 5,\quad \alpha_2 = 0,\quad \alpha_3 = -4\;\mathrm{rad/s^2}}$ The total angle covered is the area under $\omega(t)$ (a trapezium).

Links

Topics: Rotational dynamics Concepts: Angular acceleration · Torque Skills: Reading graphs Objects: Rotating disc