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Chapter 6: Problem 52

The centre of a wheel rolling on a plane surface moves with a speed \(v_{0}\). A particle on the rim of the wheel at the same level as that centre will be moving at speed \(\sqrt{n} v_{0}\) then the value of \(n\) is.

Short Answer

Expert verified
The value of \(n\) is 4.

Step by step solution

01

Identify the Knowns

Firstly, we identify the knowns and unknowns in the problem. The speed of the wheel's center, denoted as \(v_{0}\), is known. The combined speed of the particle on the rim when it is at the same level as the wheel's center is given as \(\sqrt{n} v_{0}\). The unknown is \(n\), which represents a constant in the root.
02

Analyze the Particle's Movements

With the wheel rolling, the center of the wheel is moving at speed \(v_{0}\). The particle on the rim is moving at the same speed, \(v_{0}\), due to the translational movement. However, when the particle is at the same level as the wheel's center, its overall speed is increased due to an additional speed of \(v_{0}\) from the rotational movement.
03

Solve for the Unknown

Equating the given expression for the particle's overall speed \(\sqrt{n} v_{0}\) to the sum of the particle's speed from translation and rotation (the two \(v_{0}\)'s), we have \(2v_{0} = \sqrt{n} v_{0}\). Solving for \(n\), we find that \(n = 4\).

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Key Concepts

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

Wheel Dynamics
Understanding wheel dynamics is key to solving problems involving rotational motion. A wheel typically has two types of motion as it rolls: translational motion, where the center of the wheel moves in a straight line, and rotational motion, where the wheel spins around its own axis.
In a perfect rolling scenario, these two motions contribute to the overall motion of every point on the wheel. When considering the center of the wheel, it primarily follows translational movement along the surface. This is often at a speed denoted by a variable like \(v_0\), representing the speed at which the wheel's center progresses.
Additionally, dynamics of the wheel involve no slipping condition, meaning the point of contact with the ground momentarily stays static. This complex interaction between the different movements provides unique challenges, such as analyzing the speed of different points on the wheel.
Particle Speed
The speed of a particle on the rim of a wheel varies depending on its position relative to the center. This variation is due to both translational and rotational motion.
  • Translational Speed: Each particle on the wheel's rim moves forward with the speed \(v_0\), matching the wheel's center.
  • Rotational Speed: Particles also have a speed equivalent to the wheel's rotational motion, which is again \(v_0\) at the rim.
At any point, particularly when a particle on the rim is level with the center, its speed is the sum of these motions. That's \(2v_0\) due to both movements adding up.
This compound speed can be represented as \(\sqrt{n}v_0\), which in the given scenario turns into \(2v_0\). Solving for \(n\) results in \(n = 4\). This tells us the rim particle's speed is four times a reference related to \(v_0\).
Translational Movement
Translational movement refers to the motion of the entire body (like a wheel) moving linearly along a path. Imagine the point at the center of the wheel; it moves across a surface with a consistent speed, \(v_0\), which defines this translational movement.
Despite the spinning of the wheel, each point in the wheel's center experiences purely translational motion when considering the larger context. This motion is unchanged by the wheel's rotation, contributing to holistic dynamics governing the wheel's function.
The real complexity arises when you consider each particle on the wheel not only moves forward with this translational speed but also interacts with rotational dynamics. Understanding these elements can demystify how different points on a rotating object travel at various speeds. Translational movement ensures that while the wheel rolls, each dot on the wheel describes a cycloidal path due to combined motions, ultimately linking closely to involved wheel dynamics mechanics.

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Most popular questions from this chapter

A thick-walled hollow sphere has outer radius \(R .\) It rolls down on an inclined plane without slipping and its speed at bottom is \(v_{0}\). Now the incline is waxed so that the friction becomes zero. The sphere is observed to slide down without rolling and the speed now is \(\left(5 v_{0} / 4\right)\). The radius of gyration of the hollow sphere about the axis through its centre is \(\frac{n R}{4} .\) Then the value of \(n\) is.

The angular momentum of a particle rotating with a central force is constant due to (A) constant linear momentum. (B) zero torque. (C) constant torque. (D) constant force.

A uniform rod of mass \(m\) and length \(\ell\) is placed in gravity-free space and linear impulse \(J\) is given to rod at a distance \(x=l / 4\) from centre and perpendicular to rod. Point \(A\) is at a distance \(l / 3\) from centre. $$ \begin{array}{ll} \hline \text { Column-I } & \text { Column-I } \\ \hline \text { (A) Speed of centre of rod is } & \text { (1) Zero } \\ \text { (B) Speed of point } A \text { is } & \text { (2) } \frac{5}{2} \frac{J}{m} \\ \text { (C) Speed of upper end of rod is } & \text { (3) } \frac{J}{m} \\ \text { (D) Speed of lower end of rod is } & \text { (4) } \frac{J}{2 m} \\ \text { (5) } \frac{J}{5 m} \\ \hline \end{array} $$

A hollow cylinder, a spherical shell, a solid cylinder and a solid sphere are allowed to roll on an inclined rough surface of coefficient of friction \(\mu\) and inclination \(\theta\). The correct statements are (A) if cylindrical shell can roll on inclined plane, all other objects will also roll. (B) if all the objects are rolling and have same mass, the \(\mathrm{KE}\) of all the objects will be same at the bottom of inclined plane. (C) work done by the frictional force will be zero, if objects are rolling. (D) frictional force will be equal for all the objects, if mass is same.

If initial kinetic energy of the disc is \(K\), then the kinetic energy of the disc after it starts pure rolling will be (A) \(\frac{2}{3} K\) (B) \(\frac{1}{2} K\) (C) \(\frac{1}{3} K\) (D) \(\frac{3}{4} K\)
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