Source: High school physics (Chinese)
Problem Sets:
Problem
A flywheel with a rotational speed of 1500 r/s decelerates uniformly after braking and comes to a stop in 5.0 s. The diameter of the flywheel is 1.00 m.
- What is the angular acceleration of the flywheel?
- How many revolutions does the flywheel make from the start of braking until it stops?
- What is the angular velocity of the flywheel 2.5 s after braking begins?
- What is the linear velocity of a point on the edge of the flywheel 2.5 s after braking begins?
[Q1] $\alpha = -600\pi$ rad/s$^2$ [Q2] $N = 3750$ revolutions [Q3] $\omega_1 = 1500\pi$ rad/s [Q4] $v_1 = 750\pi$ m/s
First, convert the initial rotational frequency $f_0$ to initial angular velocity $\omega_0$. The final angular velocity is $\omega_f = 0$.
$$\omega_0 = 2\pi f_0 = 2\pi(1500 \text{ r/s}) = 3000\pi \text{ rad/s}$$[Q1] Use the rotational kinematic equation for constant angular acceleration $\alpha$.
$$\omega_f = \omega_0 + \alpha t$$ $$\alpha = \frac{\omega_f - \omega_0}{t} = \frac{0 - 3000\pi \text{ rad/s}}{5.0 \text{ s}}$$[Q2] Find the total angular displacement $\Delta\theta$ and then convert it to the number of revolutions $N$.
$$\Delta\theta = \frac{1}{2}(\omega_0 + \omega_f)t = \frac{1}{2}(3000\pi \text{ rad/s} + 0)(5.0 \text{ s}) = 7500\pi \text{ rad}$$ $$N = \frac{\Delta\theta}{2\pi} = \frac{7500\pi \text{ rad}}{2\pi \text{ rad/rev}}$$[Q3] Find the angular velocity $\omega_1$ at time $t_1 = 2.5$ s.
$$\omega_1 = \omega_0 + \alpha t_1 = 3000\pi \text{ rad/s} + (-600\pi \text{ rad/s}^2)(2.5 \text{ s})$$[Q4] Find the linear velocity $v_1$ on the edge of the flywheel at time $t_1$. The radius is $R = D/2 = 1.00 \text{ m} / 2 = 0.50 \text{ m}$.
$$v_1 = R \omega_1 = (0.50 \text{ m})(1500\pi \text{ rad/s})$$