Question – On a given rough inclined plane, a solid sphere and a hollow cylinder are rolled one by one, with the same speed. What is the ratio of the heights attained by the solid sphere and the hollow cylinder?
Options:
1. $\frac{9}{10}$
2. $\frac{3}{10}$
3. $\frac{7}{10}$
4. $\frac{6}{10}$
✅ Correct Answer: Option (3)$\frac{7}{10}$
Detailed JEE Main 2024 Physics solution on rolling motion and energy conservation. Learn how to find the ratio of heights attained by a solid sphere and a hollow cylinder on an inclined plane using energy conservation principles.
JEE Main 2024 Physics Question with Solution-Concepts Involved
1. Energy Conservation in Rolling Motion
For a rolling object, the total mechanical energy is the sum of translational kinetic energy, rotational kinetic energy, and potential energy. $$\text{Total Energy} = \text{Kinetic Energy} + \text{Rotational Energy} + \text{Potential Energy} $$
$$\frac{1}{2} mv^2 + \frac{1}{2} I \omega^2 = mgh$$
where:
- $m$ = mass of the object
- $v$ = velocity of the center of mass
- $I$ = moment of inertia of the object
- $\omega$ = angular velocity
- $h$ = height attained
Since the objects roll without slipping, we use: $$v = R\:\omega$$
where $R$ is the radius of the object.
2. Moment of Inertia for Different Shapes
The moment of inertia II depends on the shape of the rolling object.
For a Solid Sphere:
$$I = \frac{2}{5} m R^2$$
For a Hollow Cylinder:
$$I = m R^2$$
3. Applying Energy Conservation
Using the equation: $$\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = mgh$$
For the solid sphere:
$$\frac{1}{2} m v^2 + \frac{1}{2} \left(\frac{2}{5} m R^2\right) \omega^2 = mgh_1$$
Since $v = R\omega$, we substitute:
$$\frac{1}{2} m v^2 + \frac{1}{2} \times \frac{2}{5} m v^2 = mgh_1 $$
$$\left(\frac{1}{2} + \frac{1}{5}\right) m v^2 = mgh_1 $$
$$\left(\frac{5}{10} + \frac{2}{10}\right) m v^2 = mgh_1 $$
$$\frac{7}{10} m v^2 = mgh_1 $$
$$h_1 = \frac{7}{10} \frac{v^2}{g}$$
For the hollow cylinder:
$$\frac{1}{2} m v^2 + \frac{1}{2} \left(m R^2\right) \omega^2 = mgh_2 $$
$$\frac{1}{2} m v^2 + \frac{1}{2} m v^2 = mgh_2 $$
$$\left(\frac{1}{2} + \frac{1}{2}\right) m v^2 = mgh_2 $$
$$m v^2 = mgh_2 $$
$$h_2 = \frac{v^2}{g}$$
4. Finding the Ratio $\frac{h_1}{h_2}$
$$\frac{h_1}{h_2} = \frac{\frac{7}{10} v^2 / g}{v^2 / g} $$
$$\frac{h_1}{h_2} = \frac{7}{10}$$
Thus, the ratio of heights attained by the solid sphere and hollow cylinder is: $\frac{7}{10}$
✅ Correct Answer: Option (3) $\frac{7}{10}$
Key Takeaways:
✔ Rolling motion involves both translational and rotational kinetic energy.
✔ The moment of inertia affects how much energy is stored as rotational energy.
✔ Solid spheres roll more efficiently compared to hollow cylinders, hence attaining a greater height.
Do You Know?
- A solid sphere has a smaller moment of inertia, meaning more energy goes into translational motion, allowing it to reach a greater height.
- A hollow cylinder has a larger moment of inertia, meaning more energy is stored in rotational motion, leading to a lower height.
- The ratio of heights depends only on the shape and not on mass or radius of the rolling object.
Practice Questions:
- A solid cylinder and a ring roll down an incline from the same height. Which one reaches the bottom first and why?
- Two spheres of the same material but different radii roll down an inclined plane. Will they reach the same height?
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