A body moves along a circular path of the radius 5 m . The coefficient of friction between the surface of the path and the body is 0.5 . The angular velocity in rad s- 1 , with which the body should move so that it does not have to leave the path is (Take g=10 m s- 2 )

Question

A body moves along a circular path of the radius 5 m . The coefficient of friction between the surface of the path and the body is 0.5 . The angular velocity in rad s- 1 , with which the body should move so that it does not have to leave the path is (Take g=10 m s- 2 ) (A) 4 (B) 3 (C) 2 (D) 1

Tardigrade Admin · Accepted Answer

Here, μ = text0.5 r = 5 m, g = 10 m s-2 The friction force provides the centripetal force â´ ( textmv2/ textr) = μ textmg or textv2 = μ textrg or or mathrmv=√μ mathrmrg=√(0.5)(5 mathrm~m)(10 mathrm~m mathrm~s-2) = 5 text m s- 1 As textv = textr ω or ω = ( textv/ textr) = (5 text m s- 1/5 text m) = 1 text rad s- 1

Q. A body moves along a circular path of the radius $5 m$ . The coefficient of friction between the surface of the path and the body is $0.5$ . The angular velocity in $r a d s^{- 1}$ , with which the body should move so that it does not have to leave the path is (Take $g = 10 m s^{- 2}$ )

$4$

$3$

$2$

$1$

Here, $μ = 0.5$
r = 5 m, g = 10 m s^-2
The friction force provides the centripetal force
$∴ \frac{mv ^{2}}{r} = μ mg$ or $v^{2} = μ rg$
or or $v = μ rg = (0.5) (5 m) (10 m s^{- 2})$
$= 5 m s^{- 1}$
As $v = r ω$ or $ω = \frac{v}{r} = \frac{5 m s ^{- 1}}{5 m} = 1 rad s^{- 1}$

Q. A body moves along a circular path of the radius 5m . The coefficient of friction between the surface of the path and the body is 0.5 . The angular velocity in rads−1 , with which the body should move so that it does not have to leave the path is (Take g=10ms−2 )

4

3

2

1

Here, μ=0.5 r = 5 m, g = 10 m s-2 The friction force provides the centripetal force ∴rmv2​=μmg or v2=μrg or or v=μrg​=(0.5)(5 m)(10 m s−2)​ =5 m s−1 As v=rω or ω=rv​=5 m5 m s−1​=1 rad s−1

Q. A body moves along a circular path of the radius $5 m$ . The coefficient of friction between the surface of the path and the body is $0.5$ . The angular velocity in $r a d s^{- 1}$ , with which the body should move so that it does not have to leave the path is (Take $g = 10 m s^{- 2}$ )

$4$

$3$

$2$

$1$

Here, $μ = 0.5$
r = 5 m, g = 10 m s^-2
The friction force provides the centripetal force
$∴ \frac{mv ^{2}}{r} = μ mg$ or $v^{2} = μ rg$
or or $v = μ rg = (0.5) (5 m) (10 m s^{- 2})$
$= 5 m s^{- 1}$
As $v = r ω$ or $ω = \frac{v}{r} = \frac{5 m s ^{- 1}}{5 m} = 1 rad s^{- 1}$