Question

You place a penny on a turntable at a distance of 14.3 cm from the center. The coefficient of static friction between the penny and the turntable is 0.400. The turntable's angular acceleration is 2.45 rad/s². How long after you turn on the turntable will the penny begin to slide off of the turntable? [Note: assume the tangential component of friction is sufficiently small and can be ignored.]

Answer 1

Answer:

Time will be 2.13 sec

Explanation:

We have given that distance r = 14.3 cm =$=\frac{14.3}{100}=0.143m$

Coefficient of static friction $\mu =0.4$

Angular acceleration is given by $\alpha =2.45rad/sec^2$

We know that frictional force is given by $F=\mu mg$

And centrifugal force is given by $F=\frac{mv^2}{r}$

When coin just starts moving

$\mu mg=\frac{mv^2}{r}$

We know that $v=\omega r$ , here $\omega$ is angular velocity

According to first equation of motion we know that

$\omega =\omega _{i}+\alpha t$

As $\omega _i=0$

so $\omega =\alpha t$

$\mu g=\frac{\left ( r\alpha t \right )^2}{r}$

$\frac{\mu g}{r}=a^2t^2$

$t=\frac{\sqrt{\frac{\mu g}{r}}}{α}$

$t=\frac{\sqrt{\frac{0.4\times 9.8}{0.143}}}{2.45}=2.13sec$