A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights h_{sph} and h_{cyl} on the incline. The ratio \(\frac{{{{\rm{h}}_{{\rm{sph}}}}}}{{{{\rm{h}}_{{\rm{cyl}}}}}}\) is given by:

This question was previously asked in

JEE Mains Previous Paper 1 (Held On: 08 Apr 2019 Shift 2)

Option 2 : \(\frac{{14}}{{15}}\)

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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**Concept: **

**Kinetic energy:**

Kinetic energy is the energy that possessed by an object due to its motion.

The kinetic energy of the rolling object is given by the formula:

\({\rm{KE}} = \frac{1}{2}{\rm{m}}{{\rm{v}}^2} + \frac{1}{2}{\rm{I}}{{\rm{\omega }}^2}\)

**Potential energy:**

The energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors.

The formula of potential energy is

PE or U = m × g × h

where

PE or U = is the potential energy of the object

m = refers to the mass of the object in kilogram (kg)

g = is the gravitational force ms^{2}

h = height of the object in meter (m)

**Calculation:**

The kinetic energy (rolling) is equal to initial potential energy (maximum height).

KINETIC ENERGY:

The kinetic energy of the rolling object is given by the formula:

\({\rm{KE}} = \frac{1}{2}{\rm{m}}{{\rm{v}}^2} + \frac{1}{2}{\rm{I}}{{\rm{\omega }}^2}\)

For rolling solid sphere, the kinetic energy is:

\({\rm{K}}{{\rm{E}}_{\rm{S}}} = \frac{1}{2}{\rm{m}}{{\rm{v}}^2} + \frac{1}{2}\left( {\frac{2}{5}{\rm{m}}{{\rm{r}}^2}} \right){{\rm{\omega }}^2}\)

For rolling solid sphere, the kinetic energy without slip is:

\({\rm{K}}{{\rm{E}}_{\rm{S}}} = \frac{1}{2}{\rm{m}}{{\rm{v}}^2} + \frac{1}{2}\left( {\frac{2}{5}{\rm{m}}{{\rm{r}}^2}} \right)\frac{{{{\rm{v}}^2}}}{{{{\rm{r}}^2}}}{\rm{\;}}\)

For rolling solid cylinder, the kinetic energy is:

\({\rm{K}}{{\rm{E}}_{\rm{C}}} = \frac{1}{2}{\rm{m}}{{\rm{v}}^2} + \frac{1}{2}\left( {\frac{1}{2}{\rm{m}}{{\rm{r}}^2}} \right){{\rm{\omega }}^2}\)

For rolling solid cylinder, the kinetic energy without slip is:

\({\rm{K}}{{\rm{E}}_{\rm{C}}} = \frac{1}{2}{\rm{m}}{{\rm{v}}^2} + \frac{1}{2}\left( {\frac{1}{2}{\rm{m}}{{\rm{r}}^2}} \right)\frac{{{{\rm{v}}^2}}}{{{{\rm{r}}^2}}}\)

POTENTIAL ENERGY:

The potential energy when the sphere reaches maximum height is:

PE_{S }= mgh_{sph}

The potential energy when the cylinder reaches maximum height is:

PE_{c} = mgh_{cyl}

Now, the sphere on reaching maximum height, the equation becomes,

\(\frac{1}{2}{\rm{m}}{{\rm{v}}^2} + \frac{1}{2}\left( {\frac{2}{5}{\rm{m}}{{\rm{r}}^2}} \right)\frac{{{{\rm{v}}^2}}}{{{{\rm{r}}^2}}} = {\rm{mg}}{{\rm{h}}_{{\rm{sph}}}}\) ----(1)

Now, the cylinder on reaching maximum height, the equation becomes,

\(\frac{1}{2}{\rm{m}}{{\rm{v}}^2} + \frac{1}{2}\left( {\frac{1}{2}{\rm{m}}{{\rm{r}}^2}} \right)\frac{{{{\rm{v}}^2}}}{{{{\rm{r}}^2}}} = {\rm{mg}}{{\rm{h}}_{{\rm{cyl}}}}\) ----(2)

On dividing, equation (1) and (2), we get,

\(\frac{{\frac{1}{2}{\rm{m}}{{\rm{v}}^2} + \frac{1}{2}\left( {\frac{2}{5}{\rm{m}}{{\rm{r}}^2}} \right)\frac{{{{\rm{v}}^2}}}{{{{\rm{r}}^2}}}}}{{\frac{1}{2}{\rm{m}}{{\rm{v}}^2} + \frac{1}{2}\left( {\frac{1}{2}{\rm{m}}{{\rm{r}}^2}} \right)\frac{{{{\rm{v}}^2}}}{{{{\rm{r}}^2}}}}} = \frac{{{\rm{mg}}{{\rm{h}}_{{\rm{sph}}}}}}{{{\rm{mg}}{{\rm{h}}_{{\rm{cyl}}}}}}\)

On cancelling, we get,

\( \Rightarrow \frac{{{{\rm{h}}_{{\rm{sph}}}}}}{{{{\rm{h}}_{{\rm{cyl}}}}}} = \left[ {\frac{{1 + \frac{2}{5}}}{{1 + \frac{1}{2}}}} \right]\)

\( \Rightarrow \frac{{{{\rm{h}}_{{\rm{sph}}}}}}{{{{\rm{h}}_{{\rm{cyl}}}}}} = \frac{{7/5}}{{3/2}}\)

\(\therefore \frac{{{{\rm{h}}_{{\rm{sph}}}}}}{{{{\rm{h}}_{{\rm{cyl}}}}}} = \frac{{14}}{{15}}\)