Question

A small button placed on a horizontal rotating platform with diameter 0.326 m will revolve with the platform when it is brought up to a rotational speed of 41.0 rev/min , provided the button is a distance no more than 0.145 m from the axis. What is the coefficient of static friction between the button and the platform? How far from the axis can the button be placed, without slipping, if the platform rotates at 63.0rev/min?

Answer 1

Answer:

(a). The coefficient of static friction between the button and the platform is 0.31.

(b). The distance from the axis is 6.97 cm.

Explanation:

Given that,

Diameter of rotating platform = 0.326 m

Radius of rotating platform = 0.163 m

Rotation speed = 41.0 rev/min

Distance = 0.145 m

We need to calculate the coefficient of static friction between the button and the platform

Using formula of frictional force and centripetal force

$\mu mg=mr\omega^2$

$\mu=\dfrac{r\omega^2}{g}$

Put the value into the formula

$\mu=\dfrac{0.163\times(41\times\dfrac{2\pi}{60})^2}{9.8}$

$\mu=0.31$

If the angular velocity 63.0 rev/min

We need to calculate the distance from the axis

Using formula of frictional force and centripetal force

$\mu mg=mr\omega^2$

$r=\dfrac{\mu g}{\omega^2}$

Put the value into the formula

$r=\dfrac{0.31\times9.8}{(63\times\dfrac{2\pi}{60})^2}$

$r=0.0697\ m$

$r=6.97\ cm$

Hence, (a). The coefficient of static friction between the button and the platform is 0.31.

(b). The distance from the axis is 6.97 cm.