**Rolling of a
Body on an Inclined Plane**

A rigid body of radius of gyration (k) and radius (R) rolls down a plane inclined at an angle (θ) with the horizontal. When a body is placed on an inclined plane, it tries to slip down and hence a static friction (f) acts upwards. This friction provides a torque which causes the body to rotate. Let, a be the linear acceleration of centre of mass and α be the angular acceleration of the body.

From force diagram: For linear motion parallel to the plane,

mg sinθ – f = ma

For rotation around the axis through centre of mass, net torque = Iα.

fR = (mk²)α

As there is no slipping, the point of contact of the body with the plane is instantaneously at rest.

v = Rω and a = Rα

Solving the following three equation for a and f:

mg sinθ – f = ma … (1)

fR = mk²α … (2)

a = Rα … (3)

\(a=\frac{g\,\sin \theta }{\left( 1+\frac{{{k}^{2}}}{{{R}^{2}}} \right)}\) and \(f=\frac{mg\,\sin \theta }{\left( 1+\frac{{{R}^{2}}}{{{k}^{2}}} \right)}\).

We can also derive the condition for
pure rolling, to avoid slipping f ≤ μ_{s}N.

\(\frac{g\,\sin \theta }{1+\frac{{{R}^{2}}}{{{k}^{2}}}}\le {{\mu }_{s}}mg\,\sin \theta \).

\({{\mu }_{s}}\ge \frac{\tan \theta }{\left( 1+\frac{{{R}^{2}}}{{{k}^{2}}} \right)}\).

This is the condition of μ_{s}.
So, that the body rolls without slipping.