Q. 184.3( 10 Votes )

# A solid sphere ro

Answer :

Both spheres are identical.

∴ Total energy,

TE = KE + PE

Where

KE is kinetic energy

PE is potential energy

Since bodies are starting from rest, there initial KE is zero,

TE_{i} = PE_{i} = mgh

Where

m is mass of the bodies

g is acceleration due to gravity

h is height of inclined plane

On reaching bottom, the potential energy drops to zero (PE_{f} = 0)

Therefore, total energy of bodies at bottom are KE of bodies.

TE_{f} = KE_{f} = 1/2 mv_{f}^{2}+ 1/2 I ω^{2}

Where

m is mass of body

v_{f} is velocity attained by body

ω is angular velocity

I is moment of inertia

By Law of conservation of energy, we have,

TE_{i} = TE_{f}

I.e. mgh = 1/2 mv_{f}^{2}+ 1/2 I ω^{2}

= 1/2 mv_{f}^{2} + 1/2 (2mR^{2}/5) ω^{2}

V = Rω

∴ mgh = 1/2 mv_{f}^{2}+ mv_{f}^{2}/5

v_{f} = (10gh/7)^{1/2}

The final velocity is independent of angle of inclination. Therefore

both bodies will have same final speed on reaching ground.

B & C

When a body is moving on an inclined plane of inclination θ, acceleration acting on body is g (acceleration due to gravity vertically downward)

The acceleration along inclined plane is g sin(θ) component of acceleration.

Let height from which body fall be H.

Then length of inclined plane of inclination θ,

L = H/sin(θ)

We know the equation,

s = u t + 1/2 at^{2}

Where

u is initial velocity

s is distance covered

a is acceleration acting on body

t is time

For case of motion along plane,

u = 0 (since body starts from rest)

s = L = H/sin(θ)

a = g sin(θ)

t is time

Substituting the values,

H/sin(θ) = 0× t + 1/2 g sin(θ) t^{2}

∴ time for reaching ground,

t = (2H/g)^{1/2} / sin(θ)

From above equation, we can infer that time taken by a body to roll down an inclined plane is inversely proportional to sin of angle of inclination (θ).

Sine increases with increase of angle, thus time for fall decreases for greater inclination (t ∝ 1/sin(θ))

Thus greater the angle of inclination, lower the time needed for reaching ground.

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