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Two identical spheres \(A\) and \(B\) are free to move and to rotate about their centres. They are given the same impulse \(J\). The lines of action of the impulses pass through the centre of \(A\) and away from the centre of \(B\), then (1) \(A\) and \(B\) will have the same speed (2) \(B\) will have greater kinetic energy than \(A\) (3) they will have the same kinetic energy, but the linear kinetic energy of \(B\) will be less than that of \(A\) (4) the kinetic energy of \(B\) will depend on the point of impact of the impulse on \(B\)

Short Answer

Expert verified
(2) B will have greater kinetic energy than A.

Step by step solution

01

Understand Impulse and Its Effects

Impulse is defined as the product of force and the time duration for which it acts, which results in a change in momentum. In this case, both spheres receive the same impulse \(J\), meaning the change in momentum for both spheres will be the same. This ensures that their linear velocities after the impulse are initially equal if there are no rotational effects.
02

Analyze Linear and Rotational Motion

Since sphere \(A\) receives the impulse through its center, it only gains linear velocity, avoiding any rotational motion. In contrast, for sphere \(B\), the impulse does not pass through its center, causing \(B\) to experience both linear motion and rotational motion due to the torque created by the offset impulse.
03

Examine Kinetic Energy Distribution in \(A\) and \(B\)

Kinetic energy for a body consists of translational kinetic energy (\( \frac{1}{2}mv^2 \)) and rotational kinetic energy (\( \frac{1}{2}I\omega^2 \)). Sphere \(A\) only has translational kinetic energy, while sphere \(B\) has both translational and additional rotational kinetic energy. Thus, \(B\) ends up having more total kinetic energy compared to \(A\).
04

Conclusion - Analyzing the Answers

We conclude that (2) \(B\) will have greater kinetic energy than \(A\). This is because \(B\) has both translational and rotational kinetic energy compared to only translational for \(A\). Options (1), (3), and (4) are not correct as they do not comprehensively address both translational and rotational aspects of motion in \(B\).

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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 a fundamental concept in mechanics, representing the energy that an object possesses due to its motion. It is essential to understand that kinetic energy can be divided into two types: translational and rotational. For a non-rotating mass, translational kinetic energy is typically described by the formula,
  • Translational Kinetic Energy: \( \frac{1}{2}mv^2 \)
where \( m \) is the mass and \( v \) is the linear velocity of the object. However, when an object also rotates, as in the case with sphere \( B \) in the original exercise, rotational kinetic energy comes into play:
  • Rotational Kinetic Energy: \( \frac{1}{2}I\omega^2 \)
where \( I \) represents the moment of inertia and \( \omega \) is the angular velocity. Therefore, the total kinetic energy of a body like sphere \( B \) includes both translational and rotational components, leading to a greater total kinetic energy compared to sphere \( A \), which possesses only translational kinetic energy. Understanding these components helps in analyzing how energy is distributed differently in objects when influenced by linear and rotational forces.
Rotational Motion
When we talk about rotational motion, we refer to the movement of an object around a particular axis. In the original exercise, the introduction of rotational motion is directly linked to the point of application of the impulse on sphere \( B \). Unlike sphere \( A \), for which the impulse passes through its center causing only linear motion, sphere \( B \) experiences a torque.
Torque is the rotational equivalent of force, which leads to the angular acceleration of an object, contributing to its rotational motion. The formula to calculate torque \( \tau \) is given by:
  • Torque: \( \tau = r \times F \)
where \( r \) is the distance from the axis of rotation to the point where the force is applied, and \( F \) is the force applied. The presence of torque causes sphere \( B \) to rotate, increasing its total kinetic energy as it includes both rotational and translational components. This concept highlights why different points of impulse application result in distinct motion types, affecting the energy distribution in rotating objects.
Linear Momentum
Linear momentum is a crucial quantity in physics, reflecting an object's tendency to continue moving in the same direction. It is mathematically defined as the product of an object’s mass and velocity
  • Linear Momentum: \( p = mv \)
In impulse mechanics, a key relationship exists between impulse and momentum. An impulse \( J \), which is the product of a force applied over a time interval, produces a change in an object's momentum:
  • Impulse: \( J = \Delta p \)
That being said, when both spheres \( A \) and \( B \) receive the same impulse, their linear momentum changes equally based on this impulse. Consequently, the initial linear speeds will be identical, if rotational effects are ignored. However, due to different points of impulse application relative to their centers, sphere \( B \) additionally gains rotational motion. This combination of linear and rotational motion implies that while their linear momentum remains affected similarly, sphere \( B \)'s higher kinetic energy results from its rotational dynamics.

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Most popular questions from this chapter

A uniform solid sphere of radius \(r\) is rolling on a smooth horizontal surface with velocity \(V\) and angular velocity \(\omega=(V=\omega r) .\) The sphere collides with a sharp edge on the wall as shown in figure. The coefficient of friction between the sphere and the edge \(\mu=1 / 5\). Just after the collision the angular velocity of the sphere becomes equal to zero. The linear velocity of the sphere just after the collision is equal to (1) \(V\) (2) \(\frac{V}{5}\) (3) \(\frac{3 V}{5}\) (4) \(\frac{V}{6}\)

A solid sphere rolls down two different inclined planes of the same height but of different inclinations (1) the speeds will be same but time of descent will be different (2) in both cases, the speeds and time of descent will be same (3) speeds and time of descent both will be different (4) the speeds will be different but time of descent will be same

Two bodies with moments of inertia \(I_{1}\) and \(I_{2}\left(I_{1}>I_{2}\right)\) have equal angular momenta. If their kinetic energies of rotation \(\operatorname{are} E_{1}\) and \(E_{2}\), respectively, then (i) \(E_{1}=E_{2}\) (2) \(E_{1}E_{2}\) (4) \(E_{1} \geq E_{2}\)

A circular platform is mounted on a vertical frictionless axle. Its radius is \(r=2 \mathrm{~m}\) and its moment of inertia is \(I=200 \mathrm{~kg} \mathrm{~m}^{2} .\) It is initially at rest. A \(70 \mathrm{~kg}\) man stands on the edge of the platform and begins to walk along the edge at speed \(v_{0}=10 \mathrm{~ms}^{-1}\) relative to the ground. The angular velocity of the platform is (1) \(1.2 \mathrm{rad} \mathrm{s}^{-1}\) (2) \(0.4 \mathrm{rad} \mathrm{s}^{-1}\) (3) \(2.0 \mathrm{rad} \mathrm{s}^{-1}\) (4) \(0.7 \mathrm{rad} \mathrm{s}^{-1}\)

A horizontal plane supports a fixed vertical cylinder of radius \(R\) and a particle is attached to the cylinder by a horizontal thread \(A B\) as shown in figure. The particle initially rest on a horizontal plane. A horizontal velocity \(v_{0}\) is imparted to the particle, normal to the threading during subsequent motion. Point out the false statements. (1) Angular momentum of particle about \(O\) remains constant. (2) Angular momentum about \(B\) remains constant. (3) Momentum and kinetic energy both remain constant. (4) Kinetic energy remains constant.

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