A particle of mass m begins to slide down


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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?

Answered by admin @



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


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