A sphere of radius 'a' and mass 'm' along a horizontal plane with constant speed v₀. It encounters an inclined plane at angle θ and climbs upward. Assuming that it rolls without slipping how far up the sphere will travel? 

Concept: 

In this conservation of energy is used so that:  

Initial energy = final energy ⇒ (K.E. + P.E.)initial = (K.E. + P.E.)Final       -----(1)

The rotational energy of the sphere = (1/2) Iω²      -----(2)

Moment of inertia of sphere = (2/5) ma²      -----(3)

Here, v = rω     ----(4)

Calculations:

Given:

a = radius of the sphere, Mass of sphere = m, the velocity of sphere = v₀, L= length of slide surface, h = height of the inclined surface

Applying geometry to the inclined surface, h = L sinθ      

Using equations (1), (2), (3), and (4) we get:

Put initial kinetic energy = 0 then, 

\(mgh = \frac12 mv^2+\frac 12 I \omega ^2\)

\(mgh = \frac12 mv^2+\frac 12 \times \frac 25 \times ma^2 \times \frac{v_0^2}{a^2}\)

\(gh = \frac12 v_0^2+\frac 15 v_0^2\)

\(gh = \frac{7}{10} \times v_0^2\)

\(h = \frac{7}{10} \times \frac{v_0^2}{g}\)........(5)

Using equation (4) we get:

From the triangle, sinθ = h/l

Then, h = l × sinθ

Put this value in equation (5) we get:

l × sinθ = \(h = \frac{7}{10} \times \frac{v_0^2}{g}\)

\(I=\frac{7}{10} \times \frac{v_0^2}{g sin \theta }\)

Hence, Option (3) is correct.