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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

The height of a solid cylinder is four times that of its radius. It is kept vertically at time \(t=0\) on a belt which is moving in the horizontal direction with a velocity \(\mathrm{v}=2.45 \mathrm{t}^{2}\) where \(\mathrm{v}\) in \(\mathrm{m} / \mathrm{s}\) and \(t\) is in second. If the cylinder does not slip, it will topple over a time \(t=\) \(\\{\mathrm{A}\\} 1\) second \(\\{\mathrm{B}\\} 2\) second \\{C \(\\}\) \\} second \(\\{\mathrm{D}\\} 4\) second

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 uniform disc of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) rolls without slipping down a plane inclined at an angle \(\theta\) with the horizontal. The frictional force on the disc is \(\\{\mathrm{A}\\}[(\mathrm{Mg} \sin \theta) / 3]\) \(\\{\mathrm{B}\\}[(2 \mathrm{Mg} \sin \theta) / 3]\) \(\\{\mathrm{C}\\} \mathrm{Mg} \sin \theta\) \(\\{\mathrm{D}\\}\) None

One quarter sector is cut from a uniform circular disc of radius \(\mathrm{R}\). This sector has mass \(\mathrm{M}\). It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is... \(\\{\mathrm{A}\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(1 / 4) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(1 / 8) \mathrm{MR}^{2}\) \\{D \(\\} \sqrt{2} \mathrm{MR}^{2}\)

The ratio of angular momentum of the electron in the first allowed orbit to that in the second allowed orbit of hydrogen atom is ...... \(\\{\mathrm{A}\\} \sqrt{2}\) \(\\{B\\} \sqrt{(1 / 2)}\) \(\\{\mathrm{C}\\}(1 / 2) \quad\\{\mathrm{D}\\} 2\)
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