Question:
A thin smooth rod of length $L$ and mass $M$ is rotating freely with angular speed $\omega_{0}$ about an axis perpendicular to the rod and passing through its center. Two beads of mass $m$ and negligible size are at the center of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod, will be:
$\frac{M \omega_{0}}{M+m}$
$\frac{M \omega_{0}}{M+3 m}$
$\frac{M \omega_{0}}{M+6 m}$
$\frac{M \omega_{0}}{M+2 m}$
Question of from chapter.
JEE Main Previous Year 9 Apr. 2019
Correct Option: 3
Solution:
Download now India’s Best Exam Preparation App
Class 9-10, JEE & NEET
Related Questions
A rod of length $L$ has non-uniform linear mass density given by $\rho(x)=a+b\left(\frac{x}{\mathrm{~L}}\right)^{2}$, where $a$ and $b$ are constants and $0 \leq x \leq \mathrm{L}$. The value of $x$ for the centre of mass of the rod is at:
The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plale) of mass $4 \mathrm{~kg}$. (The coordinates of the same are shown in figure) are:
As shown in fig. when a spherical cavity (centred at $O$ ) of radius 1 is cut out of a uniform sphere of radius $R$ (centred at $C$ ), the centre of mass of remaining (shaded) part of sphere is at $G$, i.e on the surface of the cavity. $R$ can be determined by the equation:
Three point particles of masses $1.0 \mathrm{~kg}, 1.5 \mathrm{~kg}$ and $2.5 \mathrm{~kg}$ are placed at three corners of a right angle triangle of sides $4.0 \mathrm{~cm}, 3.0 \mathrm{~cm}$ and $5.0 \mathrm{~cm}$ as shown in the figure. The center of mass of the system is at a point:
Three particles of masses $50 \mathrm{~g}, 100 \mathrm{~g}$ and $150 \mathrm{~g}$ are placed at the vertices of an equilateral triangle of side $1 \mathrm{~m}$ (as shown in the figure). The $(x, y)$ coordinates of the centre of mass will be :
Four particles A, B, C and D with masses $m_{\mathrm{A}}=m, m_{\mathrm{B}}=$ $2 \mathrm{~m}, m_{\mathrm{C}}=3 \mathrm{~m}$ and $m_{\mathrm{D}}=4 \mathrm{~m}$ are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is :
A uniform rectangular thin sheet $\mathrm{ABCD}$ of mass $\mathrm{M}$ has length a and breadth $b$, as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be :
The position vector of the centre of mass $r_{\mathrm{cm}}$ of an asymmetric uniform bar of negligible area of crosssection as shown in figure is:
A force of $40 \mathrm{~N}$ acts on a point $\mathrm{B}$ at the end of an $\mathrm{L}$-shaped object, as shown in the figure. The angle $\theta$ that will produce maximum moment of the force about point $A$ is given by:
In a physical balance working on the principle of moments, when $5 \mathrm{mg}$ weight is placed on the left pan, the beam becomes horizontal. Both the empty pans of the balance are of equal mass. Which of the following statements is correct?