A solid cylinder of mass $20 \;kg$ rotates about its axis with angular speed $100\; rad s ^{-1}$ The radius of the cylinder is $0.25 \;m$. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?
Mass of the cylinder, $m=20 kg$
Angular speed, $\omega=100$ rad $s^{-1}$
Radius of the cylinder, $r=0.25 m$
The moment of inertia of the solid cylinder:
$I=\frac{m r^{2}}{2}$
$=\frac{1}{2} \times 20 \times(0.25)^{2}$
$=0.625 kg m ^{2}$
$\therefore$ Kinetic energy $=\frac{1}{2} I \omega^{2}$
$=\frac{1}{2} \times 6.25 \times(100)^{2}=3125 J$
$\therefore$ Angular momentum, $L=I \omega$
$=6.25 \times 100$
$=62.5 Js$
A solid circular disc of mass $50 \mathrm{~kg}$ rolls along a horizontal floor so that its center of mass has a speed of $0.4 \mathrm{~m} / \mathrm{s}$. The absolute value of work done on the disc to stop it is____ $\mathrm{J}$.
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 ........ $\sec$.
An automobile engine develops $100\ kW$ when rotating at a speed of $1800\ rev/min$. What torque does it deliver .......... $N-m$
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 particle performs uniform circular motion with an angular momentum $L.$ If the angular frequency of the particle is doubled and kinetic energy is halved, its angular momentum becomes