# What is the formula for acceleration of a body rolling down a smooth inclined plane?

Posted in ROTATIONAL MOTION | Email This Post |Consider a body of mass (M) and radius (R) rolling down (without slipping) a smooth inclined plane making an angle of inclination (Θ). When a body rolls down its Potential Energy ( resting at the top of inclined plane) is converted into the Kinetic energy of translation as well as rotation. As we know that body is rolling down a smooth inclined plane this means there will be no loss of energy due to friction. So the loss in Potential Energy is same as gain in kinetic energy.

Loss in potential Energy = Gain in kinetic energy

Mgh = ½ (M v^{2}) + ½ (I ω ^{2})

From the above fig we know that h = ℓ sin Θ, substituting this in the above equation we get,

Mg ℓ sin Θ = ½ (M v^{2}) + ½ (I ω ^{2})

Mg ℓ sin Θ = ½ (M v^{2}) + ½ (M K^{2} .(v^{2} / R ^{2}) )

Where K = radius of gyration, ω = v/R and I = M K^{2}

Mv^{2} / 2 (1 + (K^{2} / R^{2})) = Mg ℓ sin Θ

v^{2} = 2 g ℓ sin Θ / (1 + (K^{2} / R^{2})) but v^{2} = 2aℓ

So Formula for acceleration of a body rolling down a smooth inclined plane,

a = g sin Θ / (1 + (K^{2} / R^{2}))

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