Acceleration of a rolling body on an Inclined Plane | Physics Grade XI

Inclined Plane

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Acceleration of a rolling body on an Inclined Plane | Physics Grade XI

Acceleration of a rolling body on an Inclined Plane

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.

Acceleration of a rolling body on an inclined plane

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 v2 [(K2/ R2) + 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 v2 [(K2/ R2) + 1]
v2 = 2g h/ [(K2/ R2) + 1] ……... (i)
 
from figure, we have
sinθ = h/s
h = s sinθ
 
substituting h = s sinθ equation (i), we get,
v2 = 2g s sinθ/ [(K2/ R2) + 1]
 
since the body is falling, we can assume u = 0, then the equation v2 = u2 + 2as becomes v2 = 2as
2as = 2g s sinθ/ [(K/ R2) + 1]
.: a = g sinθ/ [(K/ R2) + 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 mv2 + ½ Iω2 = ½ v2 (m + I2/r2)
 
As the loss of potential energy = gain in kinetic energy,
mgh = ½ v2 (m+ I2/r2)
or, mg s sinθ = ½ v2 (m+ I2/r2)
or, v2 = 2 mg s sinθ/ (m + I/r2)    which is the expression for acceleration for a rolling body in an inclined plane in terms of mass.

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