Ring - Centre of mass, moment of inertia, radius of gyration, practice problems, FAQs

We often encounter problems where we need to find the energy or work done associated with the rim of tyres of a bicycle, a flywheel in the shape of a ring with spokes etc. Though the exact shapes of these objects are difficult to analyse, we might need to approximate it to a somewhat simple shape such as a ring. Thus it becomes important to study the properties of a ring in order to analyse the real life objects.

Have you ever noticed that in the sugar cane juicer, we move the wheel faster in order to gain a high rotational speed and then we insert the sugar cane into the crusher. This a direct application of the inertia in the flywheel. Let’s understand more about such applications in this article!

Table of contents

Centre of mass of a thin uniform circular ring
Centre of mass of semicircular ring
Moment of Inertia of a Uniform Ring
Radius of gyration of a circular ring
Practice Problems
FAQs

Centre of mass of a thin uniform circular ring

Let a circular ring be placed with the centre at the origin in the XY plane.

For a continuous body, the position of centre of mass can be found using,

Let us choose an elementary mass dm at an angular position θ.

Then x=R cos θ, y=Rsinθ, dm = μR dθ [µ = mass per unit length].

The x-coordinate of the centre of mass, X_cm may be written as:

The y-coordinate of the centre of mass, Y_cmmay be written as:

$Y_{C M} = \frac{\int_{0}^{2 π} (µ R d θ) . R s i n θ}{µ 2 π R} = \frac{µ R^{2} \int_{0}^{2 π} s i n θ d θ}{µ 2 π R} = \frac{µ {R^{2}}_{}^{}^{}}{µ (2 π R)} {[- c o s θ]}^{2 π}_{0} = 0$

Thus, the centre of mass of the uniform ring lies at the centre of the ring.

Centre of mass of semicircular ring

Consider a uniform semi-circular ring of radius R and mass M as shown in the figure.

As the ring is symmetrical about the y-axis,the mass distribution will be the same with respect to the y-axis. Thus, the centre of mass will lie along the y-axis.

The coordinates of the centre of mass: (0, y_com).

Consider a small element of length dl=Rdθ, mass dm subtending angle dθ at the centre, and located at an angle θ with the positive x-axis as shown in the figure.

Now, the mass of the element is as follows:

dm = MπR× Rdθ=Mπdθ

And y = R sin θ

The centre of mass is given as follows:

⇒ycom=0π(R sin θ)Mπdθ0πMπdθ=-MRπ[cosθ]π0M=2Rπ

Therefore, the coordinates of the centre of mass is (0, y_com)=0,2Rπ

Moment of Inertia of a Uniform Ring

The moment of inertia of a uniform ring about the centroidal axis perpendicular to the plane of the ring having mass M and radius R.

R = distance of the mass element dm from the axis of rotation.

Here λis the mass per unit length

dm=λ×dl

λ=M2πRand dl=Rdθ

dm=M2πR ×Rdθ

⇒dm=M2πdθ

Iring=R2dm

⇒Iring=R202πM2πdθ

⇒Iring=M2πR202πdθ

⇒Iring=MR2

Radius of gyration of a circular ring

The moment of inertia of a continuous body in terms of radius of gyration (k)can be given as,

I=Mk2

k= effective distance from the axis of rotation and the whole mass of the system is assumed to be concentrated at that position

Comparing the above equation with the moment of inertia of the ring,

I=MR2=Mk2

⇒k=R

Practice Problems

Q. A circular ring of radius 5 mhas moment of inertia of 100 kg m2about its natural axis. This ring is converted into a circular disc having the same moment of inertia about the disc’s natural axis. Find the radius of the circular disc.

Radius of ring r=5 m

Moment of inertia of ring Iring=100 kg m2

Iring=MR2

⇒100=M×52

⇒M=100/25=4 kg

When the ring is converted into a disc, the mass remains the same.

Mass of disc M=4 kg

Moment of inertia of the disc is also made the same as that of the ring.

Idisc=100 kg m2

We know, Idisc=MR22

⇒100=4×R22

⇒R=52 m

Therefore, the radius of the disc is 52 m.

Q. Two concentric rings are placed together on a horizontal table. One of the rings has a radius of 0.5 mand mass 1 kgand other is having a radius of 1 mand mass 2 kg. Find out the net moment of inertia of the system, about an axis perpendicular to their plane and passing through their common centre.

considering mass and radius of both rings

m1=1 kg

r1=0.5 m

m2=2 kg

r2=1 m

MOI of the first ring about the axis XX' passing through its centre:

I1=m1r12=1×0.52=0.25 kg m2

MOI of the second ring about the axis XX' passing through its centre:

I1=m2r22=2×12=2 kg m2

Net moment of inertia of the system is given by:

Inet=I1+I2=0.25+2=2.25 kg m2

Q. What will be the moment of inertia of ring about the axis passing diametrically through the ring having mass Mand radius R?

Given, MOI of a ring about the axis passing through the centre of mass and perpendicular to its plane is IZ=MR2.

MOI of ring about the axis passing along its diameter is given by IXor IY

Using perpendicular axis theorem, we have

IZ=IX+IY

But, IX=IYby symmetry

⇒IX=IY=IZ2=MR22

Q. Moment of inertia of a ring of radius Rand mass Mabout a line passing through its centre and perpendicular to its plane is I. Find out the M.O.I Itabout an axis which is a tangent and also perpendicular to the plane of the circle as shown in the figure.

Given M.O.I. about the centre, IZ=Icom=I

Radius of circle =R

As we know from parallel axis theorem

It=Icom+Md2

Where dis perpendicular/shortest distance between axes of Icomand It.

∴ It=I+MR2=2MR2 [∵ I=MR2]

FAQs:

Q. What does the radius of gyration of a thin uniform circular ring being equal to the radius of the ring signify?

A. It signifies that all of the mass of the ring can be considered being concentrated at a distance equal to the radius of the ring.

Q. If a ring is rolling down an inclined path, what will be the total kinetic energy of the ring?

A. The total kinetic energy of the ring will be the sum of translational kinetic energy and rotational kinetic energy about the centre of mass(i.e. the centre of the ring).

Q. If the mass of the ring is distributed non uniformly, what happens to the centre of mass of the ring?

A. For non uniform mass distribution, the centre of mass of the ring will no longer coincide with the geometric centre of the ring, rather it will lie somewhere between the centre and the circumference of the ring towards the heavier mass side.

Q. The centre of mass of a semicircular ring placed in the first and second quadrant with centre at origin lies on the y-axis. Will the position of the centre of mass change if half part (say from the second quadrant) of the ring is removed?

A. Yes, the position of the centre of mass changes if its half part of the ring lying in the second quadrant is removed. It will then shift to the first quadrant because now all mass is present there.