A coin placed on a rotating tum table just slips it is placed at a distance of 20cm from the centre.If the angular velocity of the turn table is doubled the coin will just slip at a distance of

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Solution

$A s f r i c t i o n a l f o r c e = μ m g = m r ω^{2} S o, r ω^{2} = c o n s \tan t, a s μ a n d a r e c o n s \tan t S o, r = \frac{c o n s \tan t}{ω^{2}} . I f ω i s d o u b l e d, t h e n r a d i u s i s r e d u c e d b y 4 t i m e s . S o, d i s \tan c e f r o m t h e r a d i u s n o w b e c o m e s 20 / 4 = 5 c m$

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A coin placed on a rotating tum table just slips it is placed at a distance of 20cm from the centre.If the angular velocity of the turn table is doubled the coin will just slip at a distance of

As frictional force = μmg=mrω2So ,rω2= constant , as μ and are constantSo , r=constant ω2.If ω is doubled , then radius is reduced by 4 times.So , distance from the radius now becomes 20/4 = 5 cm