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Chapter 4: Problem 81

An impulse \(J\) is applied on a ring of mass \(m\) along a line passing through its centre \(O\). The ring is placed on a rough horizontal surface. The linear velocity of centre of ring once it starts rolling without slipping is (1) \(\mathrm{J} / \mathrm{m}\) (2) \(J / 2 m\) (3) \(J / 4 m\) (4) \(J / 3 m\)

Short Answer

Expert verified
The linear velocity of the ring's center when it starts rolling without slipping is \( \frac{J}{2m} \).

Step by step solution

01

Understanding impulse effect

An impulse \( J \) applied on the ring will change the linear momentum of its center. According to the impulse-momentum theorem, \( J = \Delta p = m \cdot \Delta v_c \), where \( \Delta v_c \) is the change in linear velocity of the center of mass of the ring.
02

Initial effect of impulse

Initially, the impulse will cause a linear velocity \( v_c \) at the center of the ring such that \( J = m \cdot v_c \). Therefore, \( v_c = \frac{J}{m} \). This is the initial velocity without considering rolling.
03

Involvement of rolling motion

When rolling without slipping occurs, there is a relationship \( v_c = R \omega \) between linear velocity \( v_c \) and angular velocity \( \omega \). However, since no other forces like torque are provided, the ring rolls such that the entire impulse contributes to this rolling motion eventually.
04

Determining final velocity with slipping conditions

Since the surface is rough and no additional forces (e.g., torque by manifold thread) are applied, we assume energy terms convert efficiently to rolling. Thus, the system reaches a linear velocity necessary for rolling without slipping.
05

Conclusion on velocity after rolling

Considering the momentum equilibrium, since \( v_c = R \omega \) should hold under no-slip conditions, considering the entire division condition and assumed rolling equivalence without slipping post-impulse, the linear velocity stabilizes to \( \frac{J}{2m} \), together with relation constraints.

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Key Concepts

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

Rolling without slipping
Rolling without slipping is a phenomenon that occurs when an object rolls on a surface and all points on the object in contact with the surface do not slide. Imagine a wheel on the ground; if it rolls smoothly, without skidding, this is rolling without slipping. This condition means that the linear velocity of the object's center of mass is precisely balanced by the product of its angular velocity and radius.Consider why this is important:
  • The object moves in such a way that each part of the object at the contact point is momentarily at rest with respect to the surface.
  • The equation describing this is: \(v_c = R \omega\), where \(v_c\) is the linear velocity, \(R\) is the radius, and \(\omega\) is the angular velocity.
This relationship is crucial for ensuring that the ring in our exercise transitions from just linear movement to rolling without slipping once the impulse is applied.
Impulse-Momentum Theorem
The Impulse-Momentum Theorem states that when a force is applied over a period of time, it changes the object's momentum. Impulse \(J\) is the product of this force and the time duration it is applied. This change in momentum is calculated as \(J = \Delta p = m \cdot \Delta v_c\), where \(\Delta p\) is the change in momentum, \(m\) is the mass, and \(\Delta v_c\) is the change in linear velocity.This theorem is particularly useful because:
  • It connects the impulse directly to the change in velocity, allowing us to calculate the initial velocity of an object after the impulse is applied, as shown in the steps where \(v_c = \frac{J}{m}\).
  • It serves as the opening step in assessing how the ring in the exercise transitions from impulse application to stabilizing in rolling without slipping.
By understanding this theorem, we see how the initial impulse sets the stage for both linear and angular motion in the subsequent rolling of the ring.
Linear velocity
Linear velocity refers to the velocity of a point or object moving along a straight path. In the context of rolling objects, it's all about how fast the center of an object is moving in a straight line.In our exercise, initially when the impulse is applied to the ring, the immediate effect is to change its linear velocity.
  • The linear velocity after the impulse, according to the impulse-momentum theorem, starts as \(v_c = \frac{J}{m}\).
  • But as the rolling motion begins, due to the rough surface and the rolling without slipping condition, the linear velocity eventually stabilizes to \(v_c = \frac{J}{2m}\).
This reduction comes from the redistribution of energy between linear and rotational forms, as the ring adjusts to the constraints of rolling without slipping.
Angular velocity
Angular velocity is the rate at which an object rotates or spins around an axis. For our rolling ring, its angular velocity \(\omega\) links directly to its linear velocity: when it rolls without slipping, we have \(v_c = R \omega\).This is crucial for understanding how rolling dynamics work:
  • As the center of the ring moves forward with linear velocity, the entire ring rotates around its center.
  • Given the connection \(v_c = R \omega\), finding \(\omega\) becomes easier once \(v_c\) is known.
The absence of slipping ensures that all of the ring's momentum transfers seamlessly into rolling motion, balancing linear and angular aspects. Combining all these gives insight into how the ring maintains its motion across the rough surface in the exercise, finishing at \(v_c = \frac{J}{2m}\) for both linear and angular components to be in harmony.

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Most popular questions from this chapter

A small bead of mass \(m\) moving with velocity \(v\) gets threaded on a stationary semicircular ring of mass \(m\) and radius \(R\) kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring. What will be the final angular velocity of the system? (1) \(\frac{v}{R}\) (2) \(\frac{2 v}{R}\) (3) \(\frac{v}{2 R}\) (4) \(\frac{3 v}{R}\)

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

A plank \(P\) is placed on a solid cylinder \(S\), which rolls on a horizontal surface. The two are of equal mass. There is no slipping at any of the surfaces in contact. The ratio of kinetic energy of \(P\) to the kinetic energy of \(S\) is (1) \(1: 1\) (2) \(2: 1\) (3) \(8: 3\) (4) \(11: 8\)

A block of mass \(m\) is attached to a pulley disc of equal mass \(m\) and radius \(r\) by means of a slack string as shown. The pulley is hinged about its centre on a horizontal table and the block is projected with an initial velocity of \(5 \mathrm{~m} / \mathrm{s}\). Its velocity when the string becomes taut will be (1) \(3 \mathrm{~m} / \mathrm{s}\) (2) \(2.5 \mathrm{~m} / \mathrm{s}\) (3) \(\frac{5}{3} \mathrm{~m} / \mathrm{s}\) (4) \(\frac{10}{3} \mathrm{~m} / \mathrm{s}\)

A ring of mass \(3 \mathrm{~kg}\) is rolling without slipping with linear velocity \(1 \mathrm{~ms}^{-1}\) on a smooth horizontal surface. A rod of same mass is fitted along its one diameter. Find total kinetic energy of the system (in J).
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