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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 2.

Step by step solution

01

Understanding relative velocity concept

The basic concept of relative velocity is defined as the velocity of an object or particle in respect to a particular frame of reference. The particle's velocity in this case includes the velocity of the particle due to the rotation of the wheel (which is the same as the linear speed of the wheel \(v_{0}\)), and the velocity of the center of the wheel (also \(v_{0}\)).
02

Relative velocity of the particle

When the point on the rim is at the same level as the center, it is moving in the same direction as the center of the wheel. Therefore, the velocities are additive. The total velocity at this point on the rim is the vector sum of \(v_{0}\) and \(v_{0}\). By using the Pythagorean theorem, this results in \(\sqrt{2}v_{0}\).
03

Finding the value of n

In the problem, it is mentioned that the velocity of the particle is \(\sqrt{n} v_{0}\). By comparison, it can be seen that \(\sqrt{2}v_{0}=\sqrt{n} v_{0}\). To identify the numerical value of \(n\), the two quantities under the square roots should be equal: hence, \(n = 2\).

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Most popular questions from this chapter

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