**Acceleration of a rolling body on an Inclined Plane**

Consider a body if circular symmetry e.g. a sphere, disc, etc. of mass m and radius R, rolling down along a plane inclined to the horizontal at an angle θ as shown in the figure.

If v be linear velocity acquired by the rolling body on covering a distance s along the plane, it descends through a vertical height h and loses potential energy.

Potential energy lost by the body = mgh

This must obviously be equal to kinetic energy gained by the body,

.: total kinetic energy gained by the body = ½ m v

^{2}[(K^{2}/ R^{2}) + 1]As no slipping occurs, mechanical energy is conserved. So, the loss in potential energy = gain in K.E and

m. g. h = ½ m v

^{2}[(K^{2}/ R^{2}) + 1]v

^{2}= 2g h/ [(K^{2}/ R^{2}) + 1] ……... (i)from figure, we have

sinθ = h/s

h = s sinθ

substituting h = s sinθ equation (i), we get,

v

^{2}= 2g s sinθ/ [(K^{2}/ R^{2}) + 1]since the body is falling, we can assume u = 0, then the equation v

^{2}= u^{2}+ 2as becomes v^{2}= 2as2as = 2g s sinθ/ [(K

^{2 }/ R^{2}) + 1].: a = g sinθ/ [(K

^{2 }/ R^{2}) + 1]The above expression can be developed in terms of mass. The total kinetic energy of the rolling object in terms of mass;

K.E. = ½ mv

^{2}+ ½ Iω^{2}= ½ v^{2}(m + I^{2}/r^{2})As the loss of potential energy = gain in kinetic energy,

mgh = ½ v

^{2}(m+ I^{2}/r^{2})or, mg s sinθ = ½ v

^{2}(m+ I^{2}/r^{2})or, v

^{2}= 2 mg s sinθ/ (m + I/r^{2}) which is the expression for acceleration for a rolling body in an inclined plane in terms of mass.
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