First slide

Force

Question

A small object placed on a rotating horizontal turn-table just slips when it is placed at a distance of 4 cm from the axis of rotation. If the angular velocity of the turn-table is doubled then the object slips when its distance from the axis of rotation is

Easy

A

1cm
B

2 cm
C

4 cm
D

8 cm

Solution

The object will not slip if the centripetal force is equal to or greater than force of friction, i.e.,
$\frac{{mv}_{b}^{2}}{4} \geq μR or {mω}^{2} r \geq μmg$
or $ω^{2} r \geq μg or r \propto 1 / ω^{2}$
$∴ \frac{r_{2}}{r_{1}} = {(\frac{ω_{1}}{ω_{2}})}^{2} or r_{2} = r_{1} {(\frac{ω_{1}}{ω_{2}})}^{2}$
or, $r_{2} = 4 \times {(\frac{ω_{1}}{2 ω})}^{2} = 1 cm$

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