Physics Q31 – JEE-MAIN 2026

Home Physics Rotational MotionQuestion #434

An object of uniform density rolls up the curved path with the initial velocity

v_{o}

as shown in the figure. If the maximum height attained by an object is

\frac{7 v_{o}^{2}}{10 g}

(

g

= acceleration due to gravity), the object is a _______.

Clue 1 of 3: Energy Conservation

The problem involves a rolling object moving against gravity, so the initial kinetic energy (translational + rotational) is converted into gravitational potential energy at the maximum height.

🧠 Full Solution Path

Step 1: Apply Conservation of Mechanical Energy✦ Active

The initial total mechanical energy (kinetic) of the rolling object is converted into gravitational potential energy at the maximum height $h$ . The total kinetic energy of a rolling object is the sum of its translational and rotational kinetic energies.

K E_{i n i t i a l} = P E_{f i n a l}

\frac{1}{2} m v_{o}^{2} + \frac{1}{2} I ω_{o}^{2} = m g h

For rolling without slipping, the relation between linear and angular velocity is $v_{o} = R ω_{o}$ , so $ω_{o} = \frac{v_{o}}{R}$ . Also, the moment of inertia $I$ can be expressed as $I = m k^{2}$ , where $k$ is the radius of gyration. Substituting these into the energy equation:

\frac{1}{2} m v_{o}^{2} + \frac{1}{2} m k^{2} {(\frac{v_{o}}{R})}^{2} = m g h

\frac{1}{2} m v_{o}^{2} (1 + \frac{k^{2}}{R^{2}}) = m g h

Simplifying for $h$ :

h = \frac{v_{o}^{2}}{2 g} (1 + \frac{k^{2}}{R^{2}})

Step 2: Substitute Given Height and Solve for

\frac{k^{2}}{R^{2}}

○ Expand

We are given that the maximum height attained is $h = \frac{7 v_{o}^{2}}{10 g}$ . Equating this with the derived expression for $h$ :

\frac{7 v_{o}^{2}}{10 g} = \frac{v_{o}^{2}}{2 g} (1 + \frac{k^{2}}{R^{2}})

Divide both sides by $\frac{v_{o}^{2}}{2 g}$ :

\frac{7}{5} = 1 + \frac{k^{2}}{R^{2}}

\frac{k^{2}}{R^{2}} = \frac{7}{5} - 1 = \frac{7 - 5}{5} = \frac{2}{5}

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation to isolate the ratio $\frac{k^{2}}{R^{2}}$ .

Step 3: Identify the Object○ Expand

Now, we compare the calculated value of $\frac{k^{2}}{R^{2}}$ with the known values for different rigid bodies:

- For a solid cylinder or disc: $I = \frac{1}{2} m R^{2} ⟹ \frac{k^{2}}{R^{2}} = \frac{1}{2}$

- For a ring or hollow cylinder: $I = m R^{2} ⟹ \frac{k^{2}}{R^{2}} = 1$

- For a solid sphere: $I = \frac{2}{5} m R^{2} ⟹ \frac{k^{2}}{R^{2}} = \frac{2}{5}$

- For a hollow sphere: $I = \frac{2}{3} m R^{2} ⟹ \frac{k^{2}}{R^{2}} = \frac{2}{3}$

The calculated value $\frac{k^{2}}{R^{2}} = \frac{2}{5}$ matches that of a solid sphere.

💡 Teacher's Secret Hint

Recall the standard moments of inertia or radius of gyration ratios for common geometric shapes.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 31 – JEE-MAIN 2026

Clue 1 of 3: Energy Conservation

🧠 Full Solution Path