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

Question

Smart Achievers · Accepted Answer

(1)

The object will slip if centripetal force $\geq$ force of friction.

$m r ω^{2} \geq μ m g;$ $r ω^{2} \geq μ g$

$r ω^{2} \geq constant,$ $or$ $(\frac{r_{1}}{r_{2}}) = {(\frac{ω_{2}}{ω_{1}})}^{2}$

$\frac{4 cm}{r_{2}} = {(\frac{2 ω}{ω})}^{2}$

$∴$ $r_{2} = 1 cm$

Solution

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

1 1 cm

2 2 cm

3 4 cm

4 8 cm

Ans.
(1)

The object will slip if centripetal force $\geq$ force of friction.

$m r ω^{2} \geq μ m g;$ $r ω^{2} \geq μ g$

$r ω^{2} \geq constant,$ $or$ $(\frac{r_{1}}{r_{2}}) = {(\frac{ω_{2}}{ω_{1}})}^{2}$

$\frac{4 cm}{r_{2}} = {(\frac{2 ω}{ω})}^{2}$

$∴$ $r_{2} = 1 cm$

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