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Chapter 5: Problem 554

Two bodies of mass \(1 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) have position vector \((\mathrm{i} \wedge+2 \mathrm{j} \wedge+\mathrm{k} \wedge)\) and $(-3 \mathrm{i}-2 \mathrm{j}+\mathrm{k})$ respectively the centre of mass of this system has a position vector ..... \(\\{\mathrm{A}\\}-2 \mathrm{i} \wedge+2 \mathrm{k} \wedge\) \(\\{B\\}-2 i \wedge-j \wedge+k \wedge\) \(\\{C\\} 2 i \wedge-j \wedge-k \wedge\) \(\\{\mathrm{D}\\}-\mathrm{i} \wedge+\mathrm{j} \wedge+\mathrm{k} \wedge\)

Short Answer

Expert verified
The position vector of the center of mass of this system is \(-2 i \wedge-j \wedge+k \wedge\).

Step by step solution

01

Write down the given information

We have: - Masses: \(m_1 = 1 \mathrm{~kg}\) and \(m_2 = 3 \mathrm{~kg}\) - Position vectors: \(\vec{r}_1 = (\mathrm{i} \wedge+2 \mathrm{j} \wedge+\mathrm{k} \wedge)\) and \(\vec{r}_2 = (-3\mathrm{i}-2 \mathrm{j}+\mathrm{k})\)
02

Calculate the center of mass position vector

Using the formula for the center of mass position vector, we have: $$\vec{R}_{\mathrm{cm}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2}$$ Substitute the given values into the formula: $$\vec{R}_{\mathrm{cm}} = \frac{1(\mathrm{i} \wedge+2 \mathrm{j} \wedge+\mathrm{k} \wedge) + 3(-3\mathrm{i}-2 \mathrm{j}+\mathrm{k})}{1+3}$$
03

Simplify the position vector

Now, perform the necessary operations: $$\vec{R}_{\mathrm{cm}} = \frac{1\,\mathrm{i} + 2\,\mathrm{j} + \mathrm{k} - 9\,\mathrm{i} - 6\,\mathrm{j} + 3\,\mathrm{k}}{4}$$ Combine the like terms: $$\vec{R}_{\mathrm{cm}} = \frac{-8\,\mathrm{i} - 4\,\mathrm{j} + 4\,\mathrm{k}}{4}$$ Divide each term by the denominator (4): $$\vec{R}_{\mathrm{cm}} = -2\,\mathrm{i} - \mathrm{j} + \mathrm{k}$$
04

Compare to the given options

We now compare our result to the given options: A) \(-2 \mathrm{i} \wedge+2 \mathrm{k} \wedge\) B) \(-2 i \wedge-j \wedge+k \wedge\) C) \(2 i \wedge-j \wedge-k \wedge\) D) \(-\mathrm{i} \wedge+\mathrm{j} \wedge+\mathrm{k} \wedge\) Our result matches option B: $$\vec{R}_{\mathrm{cm}} = -2 i \wedge-j \wedge+k \wedge$$ So, the position vector of the center of mass of this system is given by option B.

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Most popular questions from this chapter

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

A solid cylinder of mass \(\mathrm{M}\) and \(\mathrm{R}\) is mounted on a frictionless horizontal axle so that it can freely rotate about this axis. A string of negligible mass is wrapped round the cylinder and a body of mass \(\mathrm{m}\) is hung from the string as shown in figure the mass is released from rest then The angular speed of cylinder is proportional to \(\mathrm{h}^{\mathrm{n}}\), where \(\mathrm{h}\) is the height through which mass falls, Then the value of \(n\) is \(\\{\mathrm{A}\\}\) zero \(\\{\mathrm{B}\\} 1\) \(\\{\mathrm{C}\\}(1 / 2)\) \([\mathrm{D}] 2\)

Moment of inertia of a sphere of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) is \(\mathrm{I}\). keeping mass constant if graph is plotted between \(\mathrm{I}\) and \(\mathrm{R}\) then its form would be.

From a uniform circular disc of radius \(\mathrm{R}\), a circular disc of radius \(\mathrm{R} / 6\) and having centre at a distance \(+\mathrm{R} / 2\) from the centre of the disc is removed. Determine the centre of mass of remaining portion of the disc. \(\\{\mathrm{A}\\}[(-\mathrm{R}) / 70]\) \(\\{\mathrm{B}\\}[(+\mathrm{R}) / 70]\) \(\\{\mathrm{C}\\}[(-\mathrm{R}) / 7]\) \(\\{\mathrm{D}\\}[(+\mathrm{R}) / 7]\)

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