# Q. A particle is released from the top of a fixed rough sphere. The particle leaves the surface of the sphere at `theta =` 60°. Find the work done by friction on the particle.

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## Expert Answers The figure is below. When the angle is `alpha < theta` the normal component of the weight G is

`Gn = G*cos(alpha) = m*g*cos(alpha)`

The friction force is

`Ff = mu*Gn =mu*m*g*cos(alpha)`

The length of arc circle between 0 and 60 degree (`pi/3` rad) is

`L =r*theta`

therefore the infinitesimal displacement along the circle is

`dL =r*d(alpha)` where angle `alpha` is measured in radians, and r is the radius of the sphere.

The total work done by the friction force is

`W = int_0^(pi/3)Ff*r*d(alpha) =int_0^(pi/3)mu*m*g*r*cos(alpha)d(alpha)=`

`=mu*m*g*r*sin(pi/3) =0.866*mu*m*g*r`

` `

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