Answer to Question #197176 in Physics for gibson

Question

A bicycle wheel of diameter 80 cm accelerates uniformly from rest to an angular speed of

50 rad/s in 20s.

(a) Find the angular acceleration of the wheel.

2

(b) If a small stone is stack between the grooves of wheel, find the tangential and radial

acceleration of a stone.

(c) During this time, how many revolutions has the wheel turned?

(d) If the wheel turns through 20 revolutions in coming to rest, at what rate was the angular

velocity changing?

Answer 1

(a)

$$\alpha =\dfrac{\omega-\omega_0}{t},$$ $$\alpha =\dfrac{50\ \dfrac{rad}{s}}{20\ s}=2.5\ \dfrac{rad}{s^2}.$$

(b) We can find the tangential acceleration of the stone as follows:

$$a_t=r\alpha=0.8\ m\cdot2.5\ \dfrac{rad}{s^2}=2\ \dfrac{m}{s^2}.$$

We can find the radial acceleration of the stone as follows:

$$a_r=\dfrac{v^2}{r}=\dfrac{(\omega r)^2}{r}=\omega^2r,$$ $$a_r=(50\ \dfrac{rad}{s})^2\cdot0.8\ m=2000\ \dfrac{rad}{s^2}.$$

(c) We can find the number of revolutions of the wheel during the time $t=20\ s$ from the kinematic equation:

$$\theta=\omega_0t+\dfrac{1}{2}\alpha t^2,$$ $$\theta=0+\dfrac{1}{2}\cdot2.5\ \dfrac{rad}{s^2}\cdot(20\ s)^2=500\ rad.$$ $$n=500\ rad\cdot\dfrac{1\ rev}{2\pi\ rad}=80\ rev.$$

d) Let's first find the length traveled by the wheel during 20 revolutions:

$$L=2\pi nr=2\pi\cdot20\cdot0.8\ m=100\ m.$$

Then, we can find the angle covered during deceleration:

$$\theta=\dfrac{L}{r}=\dfrac{100\ m}{0.8\ m}= 125\ rad.$$

Finally, we can find the deceleration of the wheel (or the rate of change of angular velocity) from the kinematic equation:

$$\omega^2=\omega_0^2+2\alpha\theta,$$ $$\alpha=\dfrac{\omega^2-\omega_0^2}{2\theta},$$ $$\alpha=\dfrac{0-(50\ \dfrac{rev}{s}\cdot\dfrac{2\pi\ rad}{1\ rev})^2}{2\cdot125\ rad}=-395\ \dfrac{rad}{s^2}.$$

The sign minus means that the wheel decelerates.