$A$ ring of mass $m = 1 \ kg$ and radius $R = 1.25 \ m$ is kept on a rough horizontal ground. $A$ small body of same mass $m = 1 \ kg$ is stuck to the top of the ring. When it is given a slight push forward,the ring starts rolling purely on the ground. What is the maximum speed of the centre of the ring (in $m/s$)?

  • A
    $1$
  • B
    $5$
  • C
    $2$
  • D
    $10$

Explore More

Similar Questions

The key feature of Bohr's theory of the spectrum of the hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find the quantized rotational energy of a diatomic molecule,assuming it to be rigid. The rule to be applied is Bohr's quantization condition.
$1.$ $A$ diatomic molecule has a moment of inertia $I$. By Bohr's quantization condition,its rotational energy in the $n^{\text{th}}$ level $(n=1, 2, 3, \dots)$ is:
$(A) \frac{1}{n^2}\left(\frac{h^2}{8 \pi^2 I}\right)$ $(B) \frac{1}{n}\left(\frac{h^2}{8 \pi^2 I}\right)$ $(C) n\left(\frac{h^2}{8 \pi^2 I}\right)$ $(D) n^2\left(\frac{h^2}{8 \pi^2 I}\right)$
$2.$ It is found that the excitation frequency from the ground state $(n=1)$ to the first excited state $(n=2)$ of rotation for the $CO$ molecule is close to $\frac{4}{\pi} \times 10^{11} \text{ Hz}$. Then the moment of inertia of the $CO$ molecule about its centre of mass is close to (Take $h=2 \pi \times 10^{-34} \text{ Js}$):
$(A) 2.76 \times 10^{-46} \text{ kg m}^2$ $(B) 1.87 \times 10^{-46} \text{ kg m}^2$ $(C) 4.67 \times 10^{-47} \text{ kg m}^2$ $(D) 1.17 \times 10^{-47} \text{ kg m}^2$
$3.$ In a $CO$ molecule,the distance between $C$ (mass $= 12 \text{ a.m.u.}$) and $O$ (mass $= 16 \text{ a.m.u.}$),where $1 \text{ a.m.u.} = \frac{5}{3} \times 10^{-27} \text{ kg}$,is close to:
$(A) 2.4 \times 10^{-10} \text{ m}$ $(B) 1.9 \times 10^{-10} \text{ m}$ $(C) 1.3 \times 10^{-10} \text{ m}$ $(D) 4.4 \times 10^{-11} \text{ m}$
Give the answer for questions $1, 2,$ and $3$.

$A$ mass $M = 40 \ kg$ is fixed at the very edge of a long plank of mass $80 \ kg$ and length $1 \ m$ which is pivoted such that it is in equilibrium. How far (approx.) from the pivot should a mass of $100 \ kg$ be attached so that the plank starts rotating with an angular acceleration of $1 \ rad/s^2$?

Difficult
View Solution

In the following problems,indicate the correct direction of the friction force acting on the cylinder,which is pulled on a rough surface by a constant force $F$. $A$ cylinder is pulled horizontally by a force $F$ acting at a point above the centre of mass of the cylinder,as shown in the figure. The friction force can be given by which of the following diagrams?

$A$,$B$,and $C$ are a disc,a solid sphere,and a spherical shell respectively,with the same radii $(R)$ and masses $(M)$. These bodies are placed as shown in the figure. The moment of inertia of the given system about the axis $PQ$ is $\frac{x}{15} I$,where $I$ is the moment of inertia of the disc about its diameter. The value of $x$ is . . . . . . .

$A$ uniform rod is fixed to a rotating turntable so that its lower end is on the axis of the turntable and it makes an angle of $20^o$ to the vertical. (The rod is thus rotating with uniform angular velocity about a vertical axis passing through one end.) If the turntable is rotating clockwise as seen from above,what is the direction of the rod's angular momentum vector (calculated about its lower end)?

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo