Question

Derive an expression for the acceleration of a body rolling down an inclined plane.

Answer 1

Answer:

Loss in potential Energy = Gain in kinetic energy

Mgh = ½ (M v2) + ½ (I ω 2)

formula for acceleration of body rolling down a smooth inclined plane

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 v2) + ½ (I ω 2)

Mg ℓ sin Θ = ½ (M v2) + ½ (M K2 .(v2 / R 2) )

Where K = radius of gyration, ω = v/R and I = M K2

Mv2 / 2 (1 + (K2 / R2)) = Mg ℓ sin Θ

v2 = 2 g ℓ sin Θ / (1 + (K2 / R2)) but v2 = 2aℓ

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

a = g sin Θ / (1 + (K2 / R2))

Answer 2

The acceleration of a body rolling down an inclined plane is

a = gsinθ/[(K²/R²) + 1]

Given :

A body rolling down an inclined plane with acceleration 'a'

To find :

An expression for the acceleration of a body rolling down an inclined plane.

Solution :

Let 'v' be the velocity of the rolling body

's' be the distance covered

'h' be the vertical height

The loss of potential energy = mgh

Kinetic energy gained = 1/2 mv² [ (K²/R²) +1 ]

We know that,

Potential energy lost = Kinetic energy gained

Therefore,

mgh =  1/2 mv² [ (K²/R²) +1 ]

  v² = 2gh/ [ (K²/R²) +1 ]   {equation 1}

From the diagram, we know that,

sinθ = h/s

    h= s(sinθ)  {equation 2}

Substituting equation 2 in 1

v² = 2gs(sinθ) / (K²/R²) +1

As it is freely rolling down, u = 0

So, v² = 2as {v²-u² = 2as}

From this,

2as = 2gs(sinθ) / (K²/R²) +1

    a = g(sinθ) / (K²/R²) +1

Therefore, The acceleration of a body rolling down an inclined plane is

a = gsinθ/[(K²/R²) + 1]

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