/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 A circular disc of radius \(R\) ... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 78

A circular disc of radius \(R\) is removed from a bigger circular disc of radius \(2 R\) such that the circumferences of the discs coincide. The centre of mass of the new disc is \(a / R\) from the centre of the bigger disc. The value of \(a\) is (A) \(\frac{1}{2}\) (B) \(\frac{1}{6}\) (C) \(\frac{1}{4}\) (D) \(\frac{1}{3}\)

Short Answer

Expert verified
The value of \(a\) is (D) \(\frac{1}{3}\)

Step by step solution

01

Understand the Given Information

A disc of radius \(R\) is removed from a disc of radius \(2R\) such that the circumferences of the discs coincide. This means that the bottom of the smaller disc is at the bottom of the larger disc. We need to find the center of mass of the remaining disc which is \(a/R\) from the center of the bigger disc.
02

Calculate the mass of the two discs

Let the total mass of the larger disc be \(M\), and the mass of the smaller disc is \(m\). The mass is proportional to the area for objects of the same thickness and material. Since the radius of the bigger disc is double the radius of the small disc, the bigger disc has four times the area of the small disc. Therefore, \(m= M/4\).
03

Find the centre of mass

If the x-axis is selected from the center of the bigger disc towards the center of the smaller disc, then the center of mass of the bigger disc is at x=0, the center of mass of the smaller disc is at x=R. The center of mass of the remaining disc after the smaller disc is removed is given by: \[x = \frac{(M * 0) - (m * R)}{M - m}\]
04

Substitute the given values

Substituting \(m = M/4\) into the above equation gives: \[x = \frac{- M*R/4}{(4M - M)/4} = - \frac{R}{3}\] Since the center of mass is defined as \(x = a/R\), we equate these to find \(a = -1/3\)
05

Interpret the Solution

The negative sign indicates that the center of mass of the remaining disc, after the smaller disc is removed, is toward the opposite side of the center of the smaller disc. It means the center of mass is \(1/3 R\) away from the center of the bigger disc and not towards the side of the smaller disc.

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Most popular questions from this chapter

(A) Particle moving with speed (1) Kinetic \(v_{0}\) strikes to a rod placed on energy will a smooth table and sticks change. to it. (B) A thin rod of mass \(m\) and (2) Momentum length \(\ell\) inclined at an angle is conserved. \(\theta\) with horizontal is dropped on a smooth horizontal plane without any angular velocity. Its tip does not rebound after impact. (C) A solid sphere of mass (3) Angular \(m\) and radius \(R\) is rolling momentum \(\begin{array}{ll}\text { with velocity } v_{0} \text { along a } & \text { is conserved }\end{array}\) horizontal plane. It suddenly about any encounters an obstacle. point. (D) Two cylinders of radii \(r_{1}\) and (4) Angular \(r_{2}\) rotating about their axis momentum with angular speed \(\omega_{1}\) and is conserved \(\omega_{2}\) moved closer to touch \(\quad\) just before each other keeping their and after axis parallel. Cylinders first impact only slip over each other at the about point contact point but slipping \(\quad\) of impact. ceases after some time due to friction.

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