A ring of diameter 2 m oscillates as a compo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The moment of inertia $I$ of a ring about an axis passing through its rim and perpendicular to its plane is given by the parallel axis theorem: $I

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(C) The moment of inertia $I$ of a ring about an axis passing through its rim and perpendicular to its plane is given by the parallel axis theorem: $I = I_{cm} + mr^2 = mr^2 + mr^2 = 2mr^2$.However,the problem states the axis is horizontal and passes through the rim,and the ring oscillates such that its centre moves in a plane perpendicular to the ring's plane. This implies the axis of rotation is a diameter of the ring. The moment of inertia of a ring about its diameter is $I = \frac{1}{2}mr^2$.Using the parallel axis theorem,the moment of inertia about a parallel axis through the rim is $I = I_{cm} + mr^2 = \frac{1}{2}mr^2 + mr^2 = \frac{3}{2}mr^2$.The time period of a compound pendulum is $T = 2\pi \sqrt{\frac{I}{mgh}}$,where $h$ is the distance from the pivot to the centre of mass $(h = r)$.$T = 2\pi \sqrt{\frac{3/2 mr^2}{mgr}} = 2\pi \sqrt{\frac{3r}{2g}}$.Comparing this with the time period of a simple pendulum $T = 2\pi \sqrt{\frac{L_{eq}}{g}}$,we get $L_{eq} = \frac{3}{2}r$.Given diameter $d = 2 \ m$,so radius $r = 1 \ m$.Therefore,$L_{eq} = \frac{3}{2} 	imes 1 = 1.5 \ m$.

$A$ ring of diameter $2 \ m$ oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre moves in a plane which is perpendicular to the plane of the ring. The equivalent length of the simple pendulum is .... $m$

Similar Questions

The height of a liquid column in a $U$-tube is $0.3 \ m$. If the liquid in one of the limbs is depressed and then released,the time period of the liquid column will be $....... \ s$.

$A$ particle of mass $5 \times 10^{-5} \ kg$ is placed at the lowest point of a smooth parabola $x^2 = 40y$ ($x$ and $y$ in $m$). If it is displaced slightly such that it is constrained to move along the parabola,the angular frequency of oscillation (in $rad/s$) will be approximately:

$A$ sphere of radius '$r$' is kept on a concave mirror of radius of curvature '$R$'. The arrangement is kept on a horizontal table. If the sphere is displaced from its equilibrium position and left,then it executes $S$.$H$.$M$. The period of oscillation will be ($g =$ acceleration due to gravity).

$A$ metal ball released from height $h$ makes a perfectly elastic collision with the ground. The frequency of the periodic vibratory motion is $\quad(g = \text{acceleration due to gravity})$

$A$ body of mass $0.04 \text{ kg}$ executes simple harmonic motion $(SHM)$ about $x=0$ under the influence of force $F$ as shown in the graph. The time period of the motion is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module