A particle is given an initial speed u inside a smooth spherical shell of the radius R=1 m that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is

Question

A particle is given an initial speed u inside a smooth spherical shell of the radius R=1 m that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-yiopzwrolsiv.png" /> (A) g√10 (B) g (C) g √2 (D) 3g

Tardigrade Admin · Accepted Answer

u2=5gR ∴ v2=u2-2gR =5gR-2gR=3gR <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-yiopzwronltf.png" /> Tangential acceleration at B is at=g (downwards) Centripetal acceleration at B is ac=(v2/R)=3g ∴ Total acceleration will be a=√aC2 + at2=g √10

Q. A particle is given an initial speed $u$ inside a smooth spherical shell of the radius $R = 1 m$ that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is

$g 10$

$g$

$g 2$

$3 g$

$u^{2} = 5 g R$
$∴ v^{2} = u^{2} - 2 g R$
$= 5 g R - 2 g R = 3 g R$

Tangential acceleration at $B$ is
$a_{t} = g$ (downwards)
Centripetal acceleration at $B$ is
$a_{c} = \frac{v ^{2}}{R} = 3 g$
$∴$ Total acceleration will be
$a = a_{C}^{2} + a_{t}^{2} = g 10$

Q. A particle is given an initial speed u inside a smooth spherical shell of the radius R=1m that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is

g10​

g

g2​

3g

u2=5gR ∴v2=u2−2gR =5gR−2gR=3gR Tangential acceleration at B is at​=g (downwards) Centripetal acceleration at B is ac​=Rv2​=3g ∴ Total acceleration will be a=aC2​+at2​​=g10​

Q. A particle is given an initial speed $u$ inside a smooth spherical shell of the radius $R = 1 m$ that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is

$g 10$

$g$

$g 2$

$3 g$

$u^{2} = 5 g R$
$∴ v^{2} = u^{2} - 2 g R$
$= 5 g R - 2 g R = 3 g R$

Tangential acceleration at $B$ is
$a_{t} = g$ (downwards)
Centripetal acceleration at $B$ is
$a_{c} = \frac{v ^{2}}{R} = 3 g$
$∴$ Total acceleration will be
$a = a_{C}^{2} + a_{t}^{2} = g 10$