The Question is from neet-physics rotational-motion , and it was created by the author SUBHANI

Question

A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first?

SUBHANI · Accepted Answer

## CORE INFORMATION\n✓ **Answer**: B\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Conservation of Energy and Rotational Motion\n\t**Related Topics**: Rotational Kinetic Energy, Moment of Inertia, Translational Kinetic Energy\n\t**Prerequisites**: Understanding of energy conservation, rotational motion, and moments of inertia.\n## SOLUTION PATHWAY\n**Given Information**:\n\t* Disk and sphere have the same radius.\n\t* They start at the same altitude on inclined planes of the same length.\n\t* They roll without slipping.\n**Method & Steps**:\n\t1. **Energy Conservation**: The potential energy at the top of the incline is converted into kinetic energy (both translational and rotational) at the bottom. The initial potential energy is $PE = mgh$, where $m$ is mass, $g$ is gravity, and $h$ is height. The final kinetic energy is $KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$, where $v$ is linear velocity, $I$ is moment of inertia, and $\omega$ is angular velocity.\n\t\tReasoning: Energy is conserved; initial potential energy is converted to final kinetic energy.\n\t\tFormula/Rule: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$\n\t2. **Relate Linear and Angular Velocity**: For rolling without slipping, $v = r\omega$, where $r$ is the radius. Thus, $\omega = v/r$.\n\t\tReasoning: Rolling without slipping implies a relationship between linear and angular velocity.\n\t\tFormula/Rule: $v = r\omega$\n\t3. **Substitute and Simplify**: Substitute $\omega = v/r$ into the energy conservation equation: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I(v/r)^2$. The moment of inertia for a solid sphere is $I_{sphere} = \frac{2}{5}mr^2$ and for a solid disk is $I_{disk} = \frac{1}{2}mr^2$. Substitute these moments of inertia into the equation.\n\t\tWorking: For sphere: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(v/r)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$. For disk: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(v/r)^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$\n\t\tKey Point: The mass cancels out, indicating the final velocity is independent of mass. \n\t4. **Solve for Velocity**: For sphere: $v_{sphere} = \sqrt{\frac{10gh}{7}}$. For disk: $v_{disk} = \sqrt{\frac{4gh}{3}}$\n\t\tResult: Since $\sqrt{\frac{10}{7}} > \sqrt{\frac{4}{3}}$, the sphere's velocity at the bottom is greater.\n\t\tVerification: Higher velocity means shorter time to reach the bottom.\n## OPTION ANALYSIS\n[A] :x: The disk will reach the bottom later than the sphere.\n[B] ✓: The sphere will reach the bottom first because it has a smaller moment of inertia and thus converts more potential energy into translational kinetic energy.\n[C] :x: They will not reach at the same time due to differences in their moments of inertia.\n[D] :x: The mass cancels out in the energy conservation equation, so the mass does not affect the final velocity.\n## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1.[Error 1]: Ignoring the rotational kinetic energy: This will lead to the incorrect conclusion that both reach at the same time. \n\t\t2.[Error 2]: Confusing moment of inertia for different shapes: Using the wrong moment of inertia value will lead to incorrect calculations.\n\t**Quick Tips**:\n\t\t1.[Helpful shortcut]: Remember that objects with smaller moments of inertia will have more energy converted to translational kinetic energy, making them faster.\n\t\t2.[Memory aid]: Think of the sphere as being more compact, allowing it to roll faster.

SUBHANI · Answer

Disk

SUBHANI · Answer

The correct Option is "Sphere        "

SUBHANI · Answer

Both reach at the same time

SUBHANI · Answer

Depends on their masses

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