Answer to Question #169875 in Mechanics | Relativity for Stephen Ogu

Question

A flywheel 500mm in diameter is brought uniformly from rest up to a speed of 300rpm in 20 seconds. Find the velocity and acceleration of a point on the rim 2 seconds after starting from rest?

Answer 1

The velocity of a point on the rim is

$v=\omega r$

where $\omega$ - the angular speed,

r - the radius of the wheel.

First, let convert the angular speed from rpm to rad/s:

$\omega=\frac{300\cdot 2\pi} {60} =31.4\space rad/s.$

The angular acceleration is:

$\alpha=\frac{\omega-\omega_0} {t} =\frac{31.4-0} {20} = 1.57 \space rad/s^2$

The angular speed at time t is:

$\omega=\omega_0+\alpha t$

So, the velocity is:

$v=(\omega_0+\alpha t) r=(0+1.57\cdot 2)\cdot 0.5=1.57\space m/s.$

The linear acceleration is:

$a=\omega^2 r=(\omega_0+\alpha t) ^2 r=(0+1.57\cdot2)^2\cdot 0.5=19.7\space m/s^2.$

Answer: 1.57 m/s, 19.7 m/s²