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A particle with mass m begins to slide down a fixed smooth sphere from top. What is its tangential acceleration when it breaks off the sphere
one year ago

Answers : (1)

3008 Points
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

one year ago
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