Question

A wheel of diameter 26 cm turns at 1500 rpm. How far will a point on the outer rimmove in 2.0 s?1) 3.1 m12) 41 m3) 90 m4) 180 m

Answer 1

Given,

The diameter of the wheel, d=26 cm=0.26 m

Therefore the radius of the wheel, r=d/2=0.13m

Time period, t=2.0 s

Frequency of the revolution, f=1500 rpm

1 relovution=2π radian.

Therefore the angular velocity is calculated as,

`\omega=(1500*2\pi)/(60)=157.1\text{ rad/s}`

The angular velocity and angular displacement are related as,

`\omega=(\theta)/(t)`

On rearranging the above equation and substituting the known values,

`\theta=\omega t=157.1*2.0=314.2\text{ rad}`

We can calculate how far the rim has moved by using the following relationship,

`s=r\theta`

Where 's' is the distance of the movement of the rim,

On substituting the known values in the above equation,

`s=0.13*314.2=40.8\text{ m}\approx41\text{ m}`

Therefore the correct answer is option 2, 41 m