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Chapter 5: Problem 569

A straight rod of length \(L\) has one of its ends at the origin and the other end at \(\mathrm{x}=\mathrm{L}\) If the mass per unit length of rod is given by Ax where \(A\) is constant where is its centre of mass. \(\\{\mathrm{A}\\} \mathrm{L} / 3\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} 2 \mathrm{~L} / 3\) \(\\{\mathrm{D}\\} 3 \mathrm{~L} / 4\)

Short Answer

Expert verified
The center of mass of the rod is located at \(\frac{2}{3}L\).

Step by step solution

01

Understanding the center of mass formula for continuous objects

In the case of continuous mass distribution, the center of mass coordinates can be calculated as follows. For the x-coordinate of the center of mass: \[ x_{cm} = \frac{\int xdm}{\int dm} \] Here, \(dm = Axdx\), where Ax is the mass per unit length and dx is an infinitesimal length element on the rod.
02

Calculate the mass of the rod

To find the total mass (M) of the rod, we integrate the dm function over the length of the rod: \[ M = \int_0^L dm = \int_0^L Axdx \]
03

Integrate

Integrate the mass function over the length of the rod: \[ M = A\int_0^L xdx = A\left[\frac{1}{2}x^2\right]_0^L = \frac{1}{2}AL^2\]
04

Calculate the numerator of the center of mass formula

Now we need to calculate the integral in the numerator of the center of mass formula: \[ \int_0^L xdm = \int_0^L x(Axdx) = A\int_0^L x^2dx\]
05

Integrate

Integrate the function over the length of the rod: \[ A\int_0^L x^2dx = A\left[\frac{1}{3}x^3\right]_0^L = \frac{1}{3}AL^3\]
06

Calculate the center of mass

Now we can plug our results into the center of mass formula: \[ x_{cm} = \frac{\frac{1}{3}AL^3}{\frac{1}{2}AL^2} \]
07

Simplify the result

Simplify the expression to find the x-coordinate of the center of mass: \[ x_{cm} = \frac{\frac{1}{3}AL^3}{\frac{1}{2}AL^2} = \frac{2}{3}L \] So, the center of mass of the rod is located at \(\frac{2}{3}L\). The correct answer is option C.

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Most popular questions from this chapter

Two blocks of masses \(10 \mathrm{~kg}\) an \(4 \mathrm{~kg}\) are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives velocity of \(14 \mathrm{~m} / \mathrm{s}\) to the heavier block in the direction of the lighter block. The velocity of the centre of mass is : \(\\{\mathrm{A}\\} 30 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{B}\\} 20 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{C}\\} 10 \mathrm{~m} / \mathrm{s}\) \(\\{\mathrm{D}\\} 5 \mathrm{~m} / \mathrm{s}\)

Two disc of same thickness but of different radii are made of two different materials such that their masses are same. The densities of the materials are in the ratio \(1: 3\). The moment of inertia of these disc about the respective axes passing through their centres and perpendicular to their planes will be in the ratio. \(\\{\mathrm{A}\\} 1: 3\) \\{B\\} \(3: 1\) \\{C\\} \(1: 9\) \(\\{\mathrm{D}\\} 9: 1\)

A meter stick of mass \(400 \mathrm{gm}\) is pivoted at one end and displaced through an angle 600 the increase in its P.E. is \(\overline{\\{\mathrm{A}\\} 2}\) \(\\{B\\} 3\) \(\\{\) C \(\\}\) Zero \(\\{\mathrm{D}\\} 1\)

A circular disc \(\mathrm{x}\) of radius \(\mathrm{R}\) is made from an iron plate of thickness \(t\). and another disc \(Y\) of radius \(4 R\) is made from an iron plate of thickness \(t / 4\) then the rotation between the moment of inertia \(\mathrm{I}_{\mathrm{x}}\) and \(\mathrm{I}_{\mathrm{y}}\) is \(\\{\mathrm{A}\\} \mathrm{I}_{\mathrm{y}}=64 \mathrm{I}_{\mathrm{x}}\) \(\\{B\\} I_{y}=32 I_{x}\) \(\\{\mathrm{C}\\} \mathrm{I}_{\mathrm{y}}=16 \mathrm{I}_{\mathrm{x}}\) \(\\{\mathrm{D}\\} \mathrm{I}_{\mathrm{y}}=\mathrm{I}_{\mathrm{x}}\)

A circular plate of uniform thickness has a diameter of \(56 \mathrm{~cm}\). A circular portion of diameter \(42 \mathrm{~cm} .\) is removed from tve \(\mathrm{x}\) edge of the plate. Find the position of centre of mass of the remaining portion with respect to centre of mass of whole plate. $\\{\mathrm{A}\\}-7 \mathrm{~cm} \quad\\{\mathrm{~B}\\}+9 \mathrm{~cm} \quad\\{\mathrm{C}\\}-9 \mathrm{~cm} \quad\\{\mathrm{D}\\}+7 \mathrm{~cm}$
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