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Standard IX

Physics

Derivation of Position-Time Relation by Graphical Method

A small disc ...

Question

A small disc is lying on the top of a hemispherical bowl of radius $R$ . The minimum speed to be imparted to the disc, so that it may leave the bowl without slipping is

V = \sqrt{\frac{g R}{2}}

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V = 2 \sqrt{g R}

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V = \sqrt{g R}

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V = \sqrt{2 g R}

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Solution

The correct option is B $V = \sqrt{g R}$
Here, centripetal force on disc $=$ weight of the disc.
So, $\frac{m v^{2}}{R} = m g$
$\Rightarrow v = \sqrt{g R}$

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A small disc is lying on the top of a hemispherical bowl of radius R. The minimum speed to be imparted to the disc, so that it may leave the bowl without slipping is

The correct option is B V=√gRHere, centripetal force on disc = weight of the disc.So, mv2R=mg⇒v=√gR

A small disc is lying on the top of a hemispherical bowl of radius $R$ . The minimum speed to be imparted to the disc, so that it may leave the bowl without slipping is

The correct option is B $V = \sqrt{g R}$
Here, centripetal force on disc $=$ weight of the disc.
So, $\frac{m v^{2}}{R} = m g$
$\Rightarrow v = \sqrt{g R}$