A small block of mass textm is released from rest from point textA inside a smooth hemisphere bowl of radius textR , which is fixed on ground such that textOA is horizontal. The ratio text(x) of magnitude of centripetal force and normal reaction on the block at any point textB varies with θ as:

Question

A small block of mass textm is released from rest from point textA inside a smooth hemisphere bowl of radius textR , which is fixed on ground such that textOA is horizontal. The ratio text(x) of magnitude of centripetal force and normal reaction on the block at any point textB varies with θ as: Question

(A)

(B)

(C)

(D)

Tardigrade Admin · Accepted Answer

(m v2/R)=N-mgsin θ N=(m v2/R)+mgsin θ <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-cyvtva1mts4s.png" /> By energy conservation, mgR sinθ =(1/2)mv2 (m v2/R)=2mgsin θ N=3mgsin θ Ratio=(m v2/R N)=(2/3) (constant) x=(2/3)

Q. A small block of mass $m$ is released from rest from point $A$ inside a smooth hemisphere bowl of radius $R$ , which is fixed on ground such that $OA$ is horizontal. The ratio $(x)$ of magnitude of centripetal force and normal reaction on the block at any point $B$ varies with $θ$ as:

$\frac{m v ^{2}}{R} = N - m g s in θ$
$N = \frac{m v ^{2}}{R} + m g s in θ$

By energy conservation,
$m g R s in θ = \frac{1}{2} m v^{2}$
$\frac{m v ^{2}}{R} = 2 m g s in θ$
$N = 3 m g s in θ$
$R a t i o = \frac{m v ^{2}}{RN} = \frac{2}{3}$ (constant)
$x = \frac{2}{3}$

Q. A small block of mass m is released from rest from point A inside a smooth hemisphere bowl of radius R , which is fixed on ground such that OA is horizontal. The ratio (x) of magnitude of centripetal force and normal reaction on the block at any point B varies with θ as:

Rmv2​=N−mgsinθ N=Rmv2​+mgsinθ By energy conservation, mgRsinθ=21​mv2 Rmv2​=2mgsinθ N=3mgsinθ Ratio=RNmv2​=32​ (constant) x=32​

Q. A small block of mass $m$ is released from rest from point $A$ inside a smooth hemisphere bowl of radius $R$ , which is fixed on ground such that $OA$ is horizontal. The ratio $(x)$ of magnitude of centripetal force and normal reaction on the block at any point $B$ varies with $θ$ as:

$\frac{m v ^{2}}{R} = N - m g s in θ$
$N = \frac{m v ^{2}}{R} + m g s in θ$

By energy conservation,
$m g R s in θ = \frac{1}{2} m v^{2}$
$\frac{m v ^{2}}{R} = 2 m g s in θ$
$N = 3 m g s in θ$
$R a t i o = \frac{m v ^{2}}{RN} = \frac{2}{3}$ (constant)
$x = \frac{2}{3}$