A coin placed on a rotating table just slips if it is placed at a distance 4r from the centre. If we double the angular velocity of the table, then the coin will just slip when it is away from the centre at a distance equal to

Question

A coin placed on a rotating table just slips if it is placed at a distance 4r from the centre. If we double the angular velocity of the table, then the coin will just slip when it is away from the centre at a distance equal to (A) 4r (B) 2r (C) r (D) r / 4

Tardigrade Admin · Accepted Answer

∴ ω =√(Âµ g/r) ∴ ω propto (1/√r) ⇒ r propto (1/ω 2) Since ω is double thus, râ²=(4 r/4)=r

Q. A coin placed on a rotating table just slips if it is placed at a distance $4 r$ from the centre. If we double the angular velocity of the table, then the coin will just slip when it is away from the centre at a distance equal to

$4 r$

$2 r$

$r$

$r /4$

$∴ ω = \frac{µ g}{r} ∴ ω \propto \frac{1}{r} \Rightarrow r \propto \frac{1}{ω ^{2}}$
Since $ω$ is double thus,
$r' = \frac{4 r}{4} = r$

Q. A coin placed on a rotating table just slips if it is placed at a distance 4r from the centre. If we double the angular velocity of the table, then the coin will just slip when it is away from the centre at a distance equal to

4r

2r

r

r/4

∴ω=rµg​​∴ω∝r​1​⇒r∝ω21​ Since ω is double thus, r′=44r​=r

Q. A coin placed on a rotating table just slips if it is placed at a distance $4 r$ from the centre. If we double the angular velocity of the table, then the coin will just slip when it is away from the centre at a distance equal to

$4 r$

$2 r$

$r$

$r /4$

$∴ ω = \frac{µ g}{r} ∴ ω \propto \frac{1}{r} \Rightarrow r \propto \frac{1}{ω ^{2}}$
Since $ω$ is double thus,
$r' = \frac{4 r}{4} = r$