Question

A particle of mass m begins to slide down on a fixed smooth sphere from the top, when it breaks off the sphere, the tangential acceleration will be?

Answer 1

if a particle starts from the top

the speed of the particle will be given by energy conservation law

it is given as

change in the potential energy = gain in kinetic energy

$mgR(1 - cos\theta) = \frac{1}{2} mv^2$

from force equation now

since it break off at this position so normal force will be zero

$mgcos\theta = \frac{mv^2}{R}$

now we will use the value from above two equations

$mgcos\theta = 2mg*(1 - cos\theta)$

by solving above equation

$3cos\theta = 2$

$cos\theta = \frac{2}{3}$

now for tangential acceleration as we know that

$a_t = \frac{F_t}{m}$

$a_t = \frac{mgsin\theta}{m}$

$a_t = gsin\theta$

from above equation

$sin\theta = \frac{\sqrt{5}}{3}$

so tangential acceleration will be

$a_t = \frac{\sqrt{5}}{3}*g$

$a_t = 7.45m/s^2$

Answer 2

Explanation:

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