Two point masses of $0.3\ kg$ and $0.7\ kg$ are fixed at the ends of a rod of length $1.4\ m$ and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum is located at a distance of
Two point masses of $0.3\ kg$ and $0.7\ kg$ are fixed at the ends of a rod of length $1.4\ m$ and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum is located at a distance of
- [IIT 1995]
- [AIIMS 2000]
- A
$0.4\ m$ from mass of $ 0.3\ kg$
- B
$0.98\ m$ from mass of $0.3\ kg$
- C
$0.70\ m$ from mass of $0.7\ kg$
- D
$0.98\ m$ from mass of $0.7\ kg$
Similar Questions
A rod of length $50\,cm$ is pivoted at one end. It is raised such that if makes an angle of $30^o$ fro the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $rad\,s^{-1}$ ) will be $(g = 10\,ms^{-2})$
A rod of length $50\,cm$ is pivoted at one end. It is raised such that if makes an angle of $30^o$ fro the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $rad\,s^{-1}$ ) will be $(g = 10\,ms^{-2})$
- [JEE MAIN 2019]
A thin uniform rod of length $l$ and mass $m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega $. Its centre of mass rises to a maximum height of
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Explain the construction and working of an ideal lever and also explain the principle of momen of force.
Explain the construction and working of an ideal lever and also explain the principle of momen of force.
A thin and uniform rod of mass $M$ and length $L$ is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement($s$) is/are correct, when the rod makes an angle $60^{\circ}$ with vertical ? [ $g$ is the acceleration due to gravity]
$(1)$ The radial acceleration of the rod's center of mass will be $\frac{3 g }{4}$
$(2)$ The angular acceleration of the rod will be $\frac{2 g }{ L }$
$(3)$ The angular speed of the rod will be $\sqrt{\frac{3 g}{2 L}}$
$(4)$ The normal reaction force from the floor on the rod will be $\frac{ Mg }{16}$
A thin and uniform rod of mass $M$ and length $L$ is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement($s$) is/are correct, when the rod makes an angle $60^{\circ}$ with vertical ? [ $g$ is the acceleration due to gravity]
$(1)$ The radial acceleration of the rod's center of mass will be $\frac{3 g }{4}$
$(2)$ The angular acceleration of the rod will be $\frac{2 g }{ L }$
$(3)$ The angular speed of the rod will be $\sqrt{\frac{3 g}{2 L}}$
$(4)$ The normal reaction force from the floor on the rod will be $\frac{ Mg }{16}$
- [IIT 2019]
The $M.I.$ of a body about the given axis is $1.2\,kg \times m^2$ and initially the body is at rest. In order to produce a rotational kinetic energy of $1500\,joule$ an angular acceleration of $25\,rad/sec^2$ must be applied about that axis for a duration of ........ $\sec$.
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