Question

Coon kept on rotating gramaphone disc just begins to slip if its centre is at a distance of 8 cm from the centre of disc. The angular velocity is then doubled. Through what distance the coin should be shifted towards the centre, so that the coin will just slip?

Answer 1

Answer:

Explanation: the coin is also rotating along the disc . the centrifugal force acting on the coin is opposed by force of friction for equilibrium F=$$m\times\dfrac{v^2}-{r}$$

=$$m\times r \times \omega^2$$

In both cases the maximum force of friction remains same

: F=$$m\times r1\times\omega^2$$

So, $$\dfrac{r2}-{r1}$$= $$\dfrac{w1^2}-{w2^2}$$

= $$\drac{1}-{4}$$

So r2=2 cm

Thus , the should be shifted through (8-2)=6cm. Towards the centre.