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A solid sphere and a hollow sphere of equal mass and radius are placed over a rough horizontal surface after rotating it about its mass centre with same angular velocity \(\omega_{0}\). Once the pure rolling starts let \(v_{1}\) and \(v_{2}\) be the linear speeds of their centres of mass. Then (1) \(v_{1}=v_{2}\) (2) \(v_{1}>v_{2}\) (3) \(v_{1}

Short Answer

Expert verified
The linear speed of the solid sphere's center of mass is less than that of the hollow sphere; hence, \(v_1 < v_2\).

Step by step solution

01

Identify System Properties

We have a solid sphere and a hollow sphere, both with the same mass and radius, placed on a rough horizontal surface. They are initially spinning about their center with angular velocity \(\omega_0\).
02

Understand Initial Conditions

Initially, both spheres have the same angular velocity \(\omega_0\) but are not in pure rolling motion. Pure rolling begins when the linear speed at the bottom of the sphere equals the surface speed, meaning no slipping occurs.
03

Determine Moment of Inertia

For a solid sphere, the moment of inertia about the mass center is \(I_s = \frac{2}{5}mr^2\), and for a hollow sphere, it is \(I_h = \frac{2}{3}mr^2\). Here, \(m\) is the mass and \(r\) is the radius.
04

Understand Conditions for Pure Rolling

For both spheres, pure rolling means the condition \(v = r\omega\) must be satisfied, where \(v\) is the linear speed of the center of mass.
05

Apply Conservation of Angular Momentum

Angular momentum conservation around the point of contact gives insights into how linear velocity \(v\) evolves when no external torques act. For a pure rolling start, compute \(v_s\) for the solid and \(v_h\) for the hollow sphere using \(I\omega = mvr\) where \(I\) is the inertia.
06

Compare Expression for Speeds

Equating angular momentum expressions: For solid sphere: \(\frac{2}{5}mr^2\omega_0 = mr\cdot v_{s}\), For hollow sphere: \(\frac{2}{3}mr^2\omega_0 = mr\cdot v_{h}\). This implies \(v_s = \frac{2}{5}r\omega_0\) and \(v_h = \frac{2}{3}r\omega_0\).
07

Analyze Linear Speed Relationship

Since \(v_s = \frac{2}{5}r\omega_0\) and \(v_h = \frac{2}{3}r\omega_0\), clearly \(\frac{2}{5} < \frac{2}{3}\), so \(v_s < v_h\). Therefore, \(v_1 < v_2\).

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Key Concepts

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

Solid Sphere Moment of Inertia
The moment of inertia is a fundamental concept when discussing rotational motion. It tells us how an object's mass is distributed about a rotation axis and how difficult it is to change the object's rotational motion. The moment of inertia for a solid sphere about its center of mass is given by the formula:\(I_s = \frac{2}{5}mr^2\).

Here, \(m\) is the mass of the sphere and \(r\) is its radius. This equation highlights that the mass distribution of a solid sphere is more concentrated towards its center, providing relatively low resistance to changes in rotational motion compared to a hollow sphere of the same mass and size.

Understanding the solid sphere's moment of inertia is crucial when analyzing problems involving rotational dynamics, as it directly affects the angular velocity and, subsequently, the linear velocity of the sphere's center of mass during rolling motion.
Hollow Sphere Moment of Inertia
Similar to a solid sphere, a hollow sphere has a unique way in which its mass is distributed around its center, affecting its moment of inertia. For a hollow sphere, the moment of inertia about its center of mass is:\(I_h = \frac{2}{3}mr^2\).

This formula shows that the mass of a hollow sphere is evenly distributed at a greater average radial distance from the axis of rotation than that of a solid sphere. Consequently, a hollow sphere possesses a larger moment of inertia, implying greater resistance to changes in its rotational motion. This higher inertia for the same mass results in different dynamic characteristics when the hollow sphere starts rolling on a surface.

When comparing a hollow sphere to a solid one, this greater inertia means it will accelerate differently under the same conditions, leading to distinct differences in rolling behavior and speed."
Angular Momentum Conservation
Angular momentum conservation is a principle stating that if no external torque acts on a system, the angular momentum remains constant. For a sphere in rolling motion, this principle helps us understand how its motion transitions from spinning to rolling.

While initially spinning without rolling, both a solid and hollow sphere conserve angular momentum about their point of contact with the surface, as long as no external torques are introduced. This principle is represented mathematically as \(I\omega = mvr\), where \(I\) is the moment of inertia, \(\omega\) the angular velocity, \(m\) the mass, \(v\) the linear velocity, and \(r\) the radius.

To find when pure rolling (without slipping) begins, we apply this conservation principle to equate the change in rotational speed to the linear motion of the spheres. From our previous equations, we can calculate the specific linear speeds that meet the pure rolling condition for both solid and hollow spheres: - **Solid Sphere**: \(v_s = \frac{2}{5}r\omega_0\)- **Hollow Sphere**: \(v_h = \frac{2}{3}r\omega_0\)
Since \(\frac{2}{5} < \frac{2}{3}\), this principle helps illustrate why the hollow sphere achieves a higher final speed compared to a solid sphere of the same mass and size. This insight is essential for understanding motion in systems where different objects undergo rotational and translational movement simultaneously.

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Most popular questions from this chapter

A thick walled hollow sphere has outer radius \(R .\) It rolls down an inclined plane without slipping and its speed at the bottom is \(v .\) If the inclined plane is frictionless and the sphere slides down without rolling, its speed at the bottom is \(5 v / 4\). What is the radius of gyration of the sphere? (1) \(\frac{R}{\sqrt{2}}\) (2) \(\frac{R}{2}\) (3) \(\frac{3 R}{4}\) (4) \(\frac{\sqrt{3} R}{4}\)

A solid cylinder of mass \(M\) and radius \(R\) is resting on a horizontal platform (which is parallel to the \(x-y\) plane) with its axis fixed along \(Y\)-axis and free to rotate about its axis. The platform is given a motion in the \(X\)-direction given by \(x=A \cos (\omega t)\). There is no slipping between the cylinder and the platform. The maximum torque acting on the cylinder during its motion is (1) \(\frac{M \omega^{2} A R}{3}\) (2) \(\frac{M \omega^{2} A R}{2}\) (3) \(\frac{2}{3} \times M \omega^{2} A R\) (4) the situation is not possible

We have four objects: a solid sphere, a hollow sphere, a ring and a disc, all of same radius. When these are released on an inclined plane, it may happen that all of them do not perform pure rolling. But from the information of pure rolling, if one object can be confirmed to be purely rolling then it can be said that rest all will perform pure rolling. This object whose pure rolling confirms pure rolling of all other objects is (1) hollow sphere (2) solid sphere (3) ring (4) disc

A thin uniform rod of mass \(m\) and length \(l\) is kept on a smooth horizontal surface such that it can move freely. At what distance from centre of rod should a particle of mass \(m\) strike on the rod such that the point \(P\) at a distance \(I / 3\) from the end of the rod is instantaneously at rest just after the elastic collision? (1) \(/ / 2\) (2) \(1 / 3\) (3) \(l / 6\) (4) \(l / 4\)

A constant external torque \(\tau\) acts for a very brief period \(\Delta t\) on a rotating system having moment of inertia \(I\), then (1) the angular momentum of the system will change by \(\tau \Delta t\) (2) the angular velocity of the system will change by \(\tau \Delta t / I\) (3) if the system was initially at rest, it will acquire rotational kinetic energy \((\tau \Delta t)^{2} / 2 I\) (4) the kinetic energy of the system will change by \((\tau \Delta t)^{2} / 2 I\)

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