## Moment of inertia of a single particle and rigid body,Radius of gyration,angular momentum

**MOMENT OF INERTIA OF A SINGLE PARTICLE**

I = mr^{2}

**MOMENT OF INERTIA OF A RIGID BODY**

Theorem on Moment of Inertia

Theorem of Parallel Axis

I = I_{cm} +Mr^{2}

Theorem of Perpendicular Axis

Moment of Inertia is independent of Mass depend on distribution of mass

Moment of inertia of some regular shaped rigid bodies

**RADIUS OF GYRATION**

Let mass of the rigid body = M

Now I = S mr^{2}

Here K is radius of gyration

RMS value of

**Case I**

A thin disc of mass M and radius R has mass per unit area where r is the distance from its center. Its moment of inertia about an axis going through its center of mass and perpendicular to its plane is

Given, surface mass density,

So, mass of the disc can be calculated by considering small element

of area 2prdr on it and then integration it for complete disc, i.e.

….(i)

Moment of inertia about the axis of the disc,

**Case II**

Two identical spherical balls of mass M and radius R each are suck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the center of the rod is Key idea This problem will be solved by applying parallel axis theorem, which states that moment of inertia of a rigid body about any axis is equals to its moment of inertia about a parallel axis through its center of mass plus the product of the mass of the body the square of the perpendicular distance between the axis.

We know that moment of inertia (MI) about the principle axis of the sphere is given by

Using parallel axis theorem, moment of inertia about the given axis

Considering both sphere at equal distance form the axis, moment of inertia due to both sphere about this axis will be

Now, moment of inertia of rod about its perpendicular bisector axis is given by

Here, given that

L = 2R

So, total moment of inertia of the system is

**Case III**

Seven identical circular planner discs, each of mass M and radius R are welded symmetrically The moment of inertia of the arrangement about an axis normal to the plane and passing through the point P is

From theorem of parallel axis,

**Case IV Cavity**

Form a uniform circular disc of radius R and 9M, a small disc of radius is removed The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through center of disc is

**Case V**

The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. what is the ratio l/R such that the moment of inertia is minimum?

MI of a solid cylinder about its perpendicular bisector of length is

For I to be minimum,

**Case:- **Moment of Inertia

Rod → mass = m, Length = L

Disc → mass = m, Radius = R

Multiple rods

Each rod of mass m and length L

Torque of Force about the Axis of Rotation

**Case 1**

A cylinder uniform rod of mass M and length *l* is pivoted at one end so that it can rotate in a vertical plane (see the figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle q with the vertical, is

As the rod rotates in vertical plane so a torque is acting on it, which is due to the vertical component of weight of rod.

Now, Torque t = force x perpendicular distance of line of action of force from axis of rotation

Again, Torque, t = Ia

Where, I = moment of inertia =

[forced and Torque frequency along axis of rotation

passing through in end]

a = angular acceleration

A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick as shown in the figure of 2N on the ring rolls it without slipping with an acceleration of 0.3 m/s^{2}. The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the rings is .

There is no slipping between ring and ground. Hence f_{2} is not maximum. But there is slipping between ring and stick. Therefore f_{1} is maximum. Now let us write the equations

Solving above four equations we get,

## Angular momentum

**ANGULAR MOMENTUM**

(ii) Angular Momentum of a Rigid Body Rotation about a Fixed Axis

Uniform Pure Rolling

* v*_{p} = *v _{Q}*

* v *- Rw = 0

* v *= Rw

If *v*p > *v*_{Q} or *v* > Rw, the motion is said to be forward slipping and if *v*p < *v*_{Q} or *v* < Rw, the motion is said the backward slipping

Accelerated Pure Rolling

*v* = Rw

a_{t} = Ra

Rolling on Rough Incline plane

*I _{solid} < I_{hollow} *or

*a*

_{solid}> a_{hollow}*t _{solid } < t_{hollow}*

**Case **

A block of mass m and a cylinder of mass 2m are released on a rough inclined plane, inclined at an angle with the horizontal. The coefficient of friction for all the contact surface is 0.5 find the accelerations of the block and the cylinder.

Assume pure rolling.

If the block and cylinder move independently on the incline, their acceleration are

So, both the bodies will move in contact

with each other with common acceleration a which we have to calculate.

Free body diagram of block is as shown.

…..(i)

….(ii)

From Eqs. (i) and (ii), we get

The free body diagram of cylinder is as shown.

And

From Eqs. (iv) and (v), we get

Using this in Eq. (iii), we get

**Case:- **Ladder → mass = m, Length = L

For equilibrium

*f* – *N* = 0

*N*_{1} – *mg* = 0

Torque about C. M. of Ladder

Substitute value of N_{1} and f

**Case:- **Topping and Sliding

For translational equilibrium

For rotational equilibrium

block has a tendency to slide before topple

block has a tendency to topple before slide

block slides down

block topples

## Straight line motion, Fixed axis rotation

Kinetic Energy of a Rigid Body Rotating About a Fixed Axis

A uniform flat disc of mass M and radius R rotates about a horizontal axis through its center with angular speed

What is its angular momentum?

A chip of mass m breaks of the disc at an instant such that the chip rises vertically above the point at which it breaks off. What is the maximum height the chip rises to from the location of its breaking? (no impulses internal force acts on the chip)

In the process, the mechanical energy of the chip remains constant.

What are the final angular momentum and energy of the disc?

**EXTRA POINTS TO REMEMBER **

In case of pure rolling on a stationary horizontal ground (when v = Rw), following points are important to note:

Distance moved by the center of mass of the rigid body in one full rotation is 2pR.

This is because

In forward slipping s > 2pR

And in backward slipping s > 2pR

The speed of a point on the circumference of the body at the instant shown in figure is *.*

This can be shown as:

From the above expression we can see that:

The path of a point on circumference is a cycloid and the distance moved by this point in one full rotation is 8R.

In the figure, the dotted line is a cycloid and the distance A_{1}A_{2}…..A5 is 8R. this can be proved

Speed of point A at this moment is,

Distance moved by it is in time dt is,

Therefore, total distance moved in one full rotation is,

On integration we get, s = 8R.

Here, K_{R} stands for rotational kinetic energy for translation kinetic energy

For example, for a disc

**Case 3**

A frame of the reference that is accelerated with respect to an inertial frame of reference is called a non –inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity w is an example of a non –inertial frame of reference. The relationship between the force F_{rot} experienced by a particle of mass m moving on the rotating disc and the force F_{in} experienced by a particle of mass m moving on the rotating disc and the force Fin experienced by the particle in an inertial frame of reference is, , where, vrot is the velocity in the rotating frame of reference and r is the position vector of the particle with respect to the center of the disc.

Now, consider a smooth slot along a diameter of a disc of radius R rotating counter –clockwise with a constant angular speed w about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the X –axis along the slot, the Y –axis perpendicular to the slot and the z –axis along the rotation axis (w = wk). A small block of mass m is gently placed in the slot at and is constrained to move only along the slot.

Force on block slot

The net reaction of the disc on the block is

Also, reaction is due to disc surface then

**Case 4 **

A thin uniform bar lies on a frictionless horizontal surface and is free to move in any way on the surface. Its mass is 0.16 kg and length is m. Two particles, each of mass 0.08 kg are moving on the same surface and towards the bar, one with a velocity 10 ms^{-1 }and the other with 6 ms^{-1} as shown in the figure. The first particle strikers the bar at point A and the other at point B. Each of A and B is at a distance of 0.5 m from the center of the bar. The particle strike the bar at the same instant of time and stick to the bar after collision.

The velocity of center of mass of the system just after impact (in ms^{-1}) is

(a) 1 (b) 2 (c) 3 (d) 4

Let v be the center of mass is at rest all the times, because of conversation of linear momentum. So, linear velocity of the system must be zero.

From law of conversation of linear momentum.

V = 4 ms^{-1}

The angular velocity of the system just after impact (in rad/s) is

(a) 8 (b) 4 (c) 2 (d) 1

AC = BC = 0.5 m

From conversation of angular momentum about C

(0.08)(10)(0.5)-(0.08)(6) (0.5)

Where,

Using this in the equation above, we get ω

**Case 5**

A light ring with three rods, each of mass m is welded on this ring. The rods form an equilateral triangle. The rigid assembly is released on a rough fixed inclined plane. Determine the minimum value of the coefficient of static friction that will allow pure rolling of the assembly.

….(i)

……(ii)

….(iii)

For no slipping …..(iv)

The moment of inertia of the assembly about its center of mass is

Now, on solving Eqs. (i), (ii), (iii) and (iv) simultaneously, we obtain

If µ is the coefficient of fricti on at the contact surface, then