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PHXI05:LAWS OF MOTION

363244 If the radius of curvature of the path of two particles of same mass are in the ratio $3 : 4$ , then in order to have constant centripetal force, their velocities will be in the ratio of

\sqrt{3} : 1

1 : \sqrt{3}

2 : \sqrt{3}

\sqrt{3} : 2

Explanation:

As per question
$F = \frac{m v^{2}}{r} =$ Constant
$\Rightarrow \frac{v^{2}}{r} =$ Constant $\Rightarrow v^{2} α r$
$\Rightarrow v α \sqrt{r}$
$∴ \frac{v_{1}}{v_{2}} = \frac{\sqrt{r_{1}}}{\sqrt{r_{2}}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

JEE - 2024

PHXI05:LAWS OF MOTION

363245 $A rod A B$ of length 2 $m$ is hinged at point $A$ and its other end $B$ is attached to a platform on which a block of mass $m$ is kept. Rod rotates about point $A$ maintaining angle $θ = 30^{\circ}$ with the vertical in such a way that platform remains horizontal and revolves on the horizontal circular path. If the coefficient of static friction between the block and platform is $μ = 0.1$ , then find the maximum angular velocity of rod so that the block does not slip on the platform. $(g = 10 m s^{- 2})$
supporting img

5 r a d s^{- 1}

8 r a d s^{- 1}

1 r a d s^{- 1}

3 r a d s^{- 1}

Explanation:

$N = m g, μ N = m r ω^{2}$
supporting img

$μ m g = m 2 \sin θ ω^{2}$
$\Rightarrow ω = \sqrt{\frac{μ g}{2 \sin θ}} = \sqrt{\frac{0.1 \times 10}{2 \times \sin 30}} = 1 r a d s^{- 1}$

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PHXI05:LAWS OF MOTION

363244 If the radius of curvature of the path of two particles of same mass are in the ratio $3 : 4$ , then in order to have constant centripetal force, their velocities will be in the ratio of

\sqrt{3} : 1

1 : \sqrt{3}

2 : \sqrt{3}

\sqrt{3} : 2

Explanation:

JEE - 2024

PHXI05:LAWS OF MOTION

5 r a d s^{- 1}

8 r a d s^{- 1}

1 r a d s^{- 1}

3 r a d s^{- 1}

Explanation:

$N = m g, μ N = m r ω^{2}$
supporting img

$μ m g = m 2 \sin θ ω^{2}$
$\Rightarrow ω = \sqrt{\frac{μ g}{2 \sin θ}} = \sqrt{\frac{0.1 \times 10}{2 \times \sin 30}} = 1 r a d s^{- 1}$

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