/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 56 A solid sphere rolls down an inc... [FREE SOLUTION] | 91Ó°ÊÓ

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A solid sphere rolls down an inclined plane and its velocity at the bottom is \(v_{1}\). Then same sphere slides down the plane (without friction) and let its velocity at the bottom be \(v_{2}\). Which of the following relation is correct (a) \(v_{1}=v_{2}\) (b) \(v_{1}=\frac{5}{7} v_{2}\) (c) \(v_{1}=\frac{7}{5} v_{2}\) (d) None of these

Short Answer

Expert verified
The correct relation is \(v_{1}=\frac{5}{7} v_{2}\), which is option (b).

Step by step solution

01

Understand the physics behind the problem

To solve this problem, we need to use the principle of conservation of energy which states that the total energy of a closed system will remain constant. For the rolling state, the total energy is part Potential Energy (PE) and part Kinetic Energy (KE), and the KE is further divided into Translational (due to motion) and Rotational (due to rotation while moving). Similarly, for the sliding state, the total energy is part PE and part Translational KE as the sphere isn't rotating.
02

Analyze the Rolling State

In the rolling state, the Kinetic Energy (KE) is represented by \(\frac{1}{2}I\omega^2\) where I is moment of inertia and \(\omega\) is angular velocity. For a solid sphere, the moment of inertia \(I=\frac{2}{5}mr^2\), where m is the mass and r the radius of the sphere, and the angular velocity is the ratio of linear velocity to radius \(\omega = \frac{v_1}{r}\). Substituting these values, KE for rolling becomes \(\frac{1}{2}\cdot \frac{2}{5}mr^2 \cdot \left(\frac{v_1}{r}\right)^2\), which simplifies to \(\frac{1}{5}mv_1^2\). This KE has to be added to the Translational KE \(\frac{1}{2}mv_1^2\) for the rolling condition so total KE rolling= \(\frac{1}{2}mv_1^2+\frac{1}{5}mv_1^2=\frac{7}{10}mv_1^2.\)
03

Analyze the Sliding State

In the sliding state, the sphere is not rotating, so the KE is entirely Translational KE which is \(\frac{1}{2}mv_2^2.\)
04

Apply Energy Conservation

As the sphere is starting from rest in both cases, total energy at the top is equal to total KE at the bottom. For the rolling case, PE (at the top)= KE (rolling at the bottom)= \(\frac{7}{10}mv_1^2.\) For the sliding case, PE (at the top)= KE(sliding at bottom)=\(\frac{1}{2}mv_2^2. \) Since the energy at the top is the same in both cases, these two equations should be equal to each other, leading to \(\frac{7}{10}mv_1^2= \frac{1}{2}mv_2^2\). Solving for \(v_1\) in terms of \(v_2\), we get \(v_1=\frac{\sqrt{10/14}}{\sqrt{2/2}}v_2\) which simplifies to \(v_1=\frac{5}{7}v_2\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is an essential concept in physics that describes the energy an object possesses due to its motion. Whenever something moves, like a rolling sphere or a sliding object, it carries kinetic energy. This energy is related to the object's mass and velocity and is expressed by the formula:
  • Translational Kinetic Energy: \( KE = \frac{1}{2}mv^2 \)
  • Rotational Kinetic Energy: \( KE = \frac{1}{2}I\omega^2 \)
In our exercise, there are two stages to consider: rolling and sliding. In the rolling stage, the sphere has both translational and rotational kinetic energy. The translational part deals with moving in a straight line, while the rotational part arises because the sphere is also spinning.
All the kinetic energy comes from the sphere's potential energy at the top of the inclined plane. As the sphere moves down, it loses potential energy but gains kinetic energy.
In contrast, during the sliding stage, the sphere's motion is linear, contributing only to translational kinetic energy, without any spinning.
Potential Energy
Potential energy represents the stored energy an object has because of its position in a force field, most commonly a gravitational field. This energy comes from the object's weight and how high it is above the ground. The mathematical expression for gravitational potential energy is:
  • Potential Energy: \( PE = mgh \)
Where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( h \) is the height.
For both rolling and sliding spheres at the top of the inclined plane, potential energy is at maximum since their initial velocity is zero. As the sphere moves down, the potential energy is converted into kinetic energy.
It's crucial to understand that while rolling, some of this potential energy changes to rotational kinetic energy as well, which doesn't occur when the sphere is sliding down without friction.
Moment of Inertia
The moment of inertia is a concept connected to rotational motion. It tells us how difficult it is to change an object's rotation around a certain axis. The higher the moment of inertia, the harder it is to make that object start or stop spinning.
For different shapes, the moment of inertia is different, and for a solid sphere, it is given by:
  • Moment of Inertia for a solid sphere: \( I = \frac{2}{5}mr^2 \)
Where \( m \) is the mass of the sphere and \( r \) is its radius.
In rolling motion, this moment of inertia has a significant role, contributing to the overall kinetic energy of the sphere. As described in the exercise, it affects the calculation of rotational kinetic energy, which combines with translational kinetic energy to give the total energy of the rolling sphere.
Only in the rolling case does this rotational kinetic energy component emerge, demonstrating how the sphere's rotational inertia influences its energy conservation at the bottom of the hill.

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