Question

A belt runs on a wheel of radius 30cm. During the time that wheel coasts uniformly to rest from an initial speed of 2.0revs/s, 25 m of belt length passes over the wheel. Find the deceleration of the wheel and the number of revolutions it turns while stopping.

Answer 1

Number of turns is given by, N = l/2πr ......(1)

The angle covered during deceleration

$\theta=2\pi N=\frac{l}{ r}$ [ from equation (1),

$\theta=\frac{25}{0.3}$ [ as given, l = 25m and r = 30cm ]

given, angular velocity, $\omega$ = 2 rev/s

The initial angular velocity, $\omega=2\pi(2)=4\pi$ rad/s

now, using formula, $\omega'^2=\omega^2+2\alpha\theta$

here, $\alpha$ angular deceleration. after revolving $\theta$ angular velocity becomes zero.

so, $\omega'=0$

now, $\alpha=\frac{\omega^2}{2\theta}$

= (4π)²/2(25/0.3)

= 16(0.3)π²/50 = 0.96π² rad/s²

The number of revolutions it turns while stopping $N=\frac{25}{2\pi (0.3)}=13$