A coin, placed on a rotating turn-table slips, when it is placed at a distance of 9 cm from the centre. If the angular velocity of the turn-table is trippled, it will just slip, if its distance from the centre is

27 cm

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9 cm

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3 cm

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1 cm

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Solution

The correct option is D 1 cm
In the given condition friction provides the required centripetal force and that is constant. i.e
m $ω^{2}$ r = constant
$\Rightarrow$ r = $α$ $\frac{1}{ω^{2}}$ $∴$ $r_{2}$ = $r_{1}$ ${(\frac{ω_{1}}{ω_{2}})}^{2}$ = 9 ${(\frac{1}{3})}^{2}$ = 1 cm

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A coin, placed on a rotating turn-table slips, when it is placed at a distance of 9 cm from the centre. If the angular velocity of the turn-table is trippled, it will just slip, if its distance from the centre is

The correct option is D 1 cmIn the given condition friction provides the required centripetal force and that is constant. i.e mω2 r = constant ⇒ r = α 1ω2 ∴ r2 = r1 (ω1ω2)2 = 9 (13)2 = 1 cm

The correct option is D 1 cm
In the given condition friction provides the required centripetal force and that is constant. i.e
m $ω^{2}$ r = constant
$\Rightarrow$ r = $α$ $\frac{1}{ω^{2}}$ $∴$ $r_{2}$ = $r_{1}$ ${(\frac{ω_{1}}{ω_{2}})}^{2}$ = 9 ${(\frac{1}{3})}^{2}$ = 1 cm