# Laws Of Rotational Motion (Part-2)

## Laws Of Rotational Motion (Part-2)

### Laws Of Rotational Motion (Part-2)

Perpendicular axis theorem:

**The moment of inertia of a plane lamina about an axis perpendicular to the plane is equal to the sum of the moments of inertia about two mutually perpendicular axis in the plane of the lamina such that the three mutually perpendicular axes have a common point of intersection.**

**Explanation:** If *I _{x}* and

*I*be the moments of inertia of a plane

_{y }lamina about perpendicular axes, OX and OY, which lie on the

plane of the lamina and intersect each other at O, (Figure Right

hand side) the moment of inertia I_{z }about an axis, say OZ,

passing through O and perpendicular to the plane of the lamina,

is given by, I_{x} + I_{y} = I_{z}

Proof:** **Let us consider a plane lamina having axes OX and OY in the plane of the lamina. The axis OZ passes through O and perpendicular to the plane of the lamina. Let the lamina is composed of large number of particles of mass *m*. Let a particle of m be at P with co-ordinates (x,y) and located at a distance r from O. Here O is the intersection of the three axes.

Now the moment of inertia of the particle P about the axis

OZ *= mr ^{2 }= m(x^{2}+y^{2}) [*

*∵*

*r*

^{2 }= x^{2}+y^{2}*]*

The moment of inertia of the whole lamina about the axis OX is given by,

*I _{x}=*

*Σ*

*my*

^{2}Similarly, the moment of inertia of the whole lamina about the axis OY is given by,

*I _{y}=*

*Σ*

*mx*

^{2}The moment of inertia of the whole lamina about OZ axis,

*I _{z}=*

*Σ*

*mr*

^{2}=*Σm(*

*x*

^{2}+y^{2}*)*

*=Σm**x ^{2}+*

*Σm*

*y*

^{2}= I_{x} + I_{y}

I_{z} = I_{x}+I_{y} So, the theorem is Proved.

**Laws Of Rotational Motion (Part-2)**

Parallel axis theorem:

**The moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through the center of mass and the product of the mass of the body and the square of perpendicular distance between the two parallel axes.**

**Explanation:** Let AB an axis on the plane of the paper and CD

Is another axis parallel to AB. Let the axis CD pass through center of

mass G of the plane lamina (Fig: Right hand side). The distance between

AB and CD axis is *h*. Now if the moment of inertia of the lamina with

respect to AB and CD axis are respectively I and I_{G}, then according to

this theorem

I=I_{G}*+*M*h ^{2}*

Proof: Let the lamina be composed of particles of mass *m _{1}, m_{2}, m_{3}* etc. and the respective distance from the axis CD are

*x*etc. Then moment of inertia of the particle of mass

_{1}, x_{2}, x_{3},*m*about axis AB

_{1 }*=*

*m*

_{1}(x_{1}+h)^{2 }= m_{1}x_{1}^{2}+m_{1}h^{2}+2m_{1}x_{1}hSimilarly, moment of inertia of the particle of mass *m _{2 }*about axis AB

*= m _{2}x_{2}^{2}+m_{2}h^{2}+2m_{2}x_{2}h*

And for* m _{3 }=*

*m*

_{3}x_{3}^{2}+m_{3}h^{2}+2m_{3}x_{3}hNow if I is the moment of inertia of the whole lamina about AB, then I will be the summation of the moment of inertia of individual particles i.e.

∴ I* = m _{1}x_{1}^{2}+m_{1}h^{2}+2m_{1}x_{1}h + m_{2}x_{2}^{2}+m_{2}h^{2}+2m_{2}x_{2}h + m_{3}x_{3}^{2}+m_{3}h^{2}+2m_{3}x_{3}h + … … …*

*= **Σ** mx ^{2}+h^{2}*

*Σm+2hΣmx*

Here*, **Σmx*= total moment of the mass of the whole lamina about axis CD. But the weight of the lamina is acting downward along the line CD through the point G. So moment of the mass of the lamina about CD is zero.

i.e., *Σmx = 0*

*Again, Σm = *M and* Σ** mx ^{2}*=I

_{G}

∴ I =IG+M*h ^{2}*+0 = I

_{G}+M

*h*

^{2}Hence the theorem is proved.

*Laws Of Rotational Motion (Part-2)*

Determination of Moment of inertia and radius of gyration of a thin uniform rod about an axis through its center and perpendicular to its length:

Let AB be a thin uniform rod of length *l*and mass M, free to rotate about the axis CD which is passing through the center and perpendicular to the length of the rod (In the figure below) The moment of inertia about the axis CD is to be found out.

Since the rod is uniform, the mass per unit length

So at distance *x* from the axis CD, let *dx* be a small length whose

mass, As *dx*is very small, we can consider all

the particles in *dx* are at same distance from CD. So moment of inertia of *dx* about axis CD

Now integrating the above equation within limits *x = – l/2* to *x = l/2 *we get moment of inertia of the entire rod.

Let K be the radius of gyration,

#### Determination of Moment of inertia and radius of gyration of a thin uniform rod about an axis through one end of the rod perpendicular to its length:

Let AB be a thin uniform rod of length *l*and mass M, free to rotate about the axis CD which is passing through one end of the rod and perpendicular to the length AB of the rod (In the figure below) The moment of inertia about the axis CD is to be found out.

Since the rod is uniform, the mass per unit length

So at distance *x* from the axis CD, let *dx*be a small length whose

mass, As *dx *is very small, we can consider all

the particles in *dx *are at same distance from CD. So moment

of inertia of *dx* about axis CD

Now integrating the above equation within limits *x = 0* to *x = l *we get moment of inertia of the entire rod.

Let K be the radius of gyration,

Determination of Moment of inertia and radius of gyration of a circular disk about an axis perpendicular to its plane passing through the center:

Let ABC be a circular disc and let *r* and *M*are respectively its radius and mass. Let O be the center of the disc and PQ its axis about which the disc rotates. The moment of inertia and radius of gyration for circular disc are to be determined.

Now, area of the disc, A = π r^{2}

∴ Mass per unit area of the disc,

Now, let us consider a small circular strip of width *dx* at a distance

*x *from the center of the disc. (Fig right hand side)

The area of the strip,

*dA**= *circumference of the strip × width of the strip

*=2π**xdx*

If *dm *is the mass of the strip, then

Now the moment of inertia of this strip about the axis PQ,

If we consider that the whole disc is composed of such like strips, then the moment for the whole disc can be obtained by integrating the above equation for the limits *x = 0* to *x = r*.

So moment of inertia,

Let K be the radius of gyration,

Determination of Moment of inertia and radius of gyration of a rectangular lamina about an axis perpendicular to its center of mass:

Let, ABCD be a rectangular lamina of mass M and length, *l *= AB = CD and breadth, b = AD = BC (Fig right hand side). Let the axis of rotation XOY pass

through the center and perpendicular to the lamina. Moment of inertia

and radius of gyration are to be determined.

Now, the axis EF passing through O is parallel to side AB and

axis GH passing through O is parallel to side AD. EF and GH are

perpendicular to each other. XOY axis is perpendicular to both EF

and GH.

According to perpendicular axis theorem, the moment of inertia of a lamina about the axis XOY,

Again, radius of gyration K is related to I as

MK^{2}=I

Expression for centripetal acceleration and centripetal force:

Let us consider an object of mass *m* is moving along a circular path of radius *r,* in anticlockwise direction with a uniform speed *v,* and an angular velocity with at center O. When the body is come from A to B in small time variation *t*. The velocity at A is the Tangent along AC.

Now, in rectangular OAEB, we get

∠AEB+∠AOB =180°

Again, ∠AEB+∠BEC = 180°

The horizontal component of velocity at A, *v _{y }= 0*

and the vertical component, *v _{x }= v*

The vertical component of velocity along AC, *v _{y }= vsinθ*

and the horizontal component, *v _{x }= vcosθ*

when *t*is very small then θ is also very small

*∴* *sin **θ = θ*, and *cos θ = 1*

*∴* the vertical component of velocity at B, *v _{y }= vθ*

and the horizontal component, *v _{x }= v*

It appears that, there is no change of velocity along horizontally,

So, centripetal acceleration,

Motion of a cyclist along a curved path:

A cyclist travelling along a curved path bends his body with the cycle

Towards center of curvature. Without centripetal force the force acting

On the cyclist would have thrown him out of the road. To balance this

tendency of skidding the cyclist along with the cycle bends his body

towards the center. This is haw centripetal force is produced.

Let a cyclist moving at uniform speed *v*on the circular path of radius

*r*bends away from the normal by an angle θ.

Let the weight of the cyclist along with the cycle be *mg *and the

Reaction force by the road on the cycle be R. R can be resolved in to two

rectangular component Rcosθ vertical to the ground and Rsinθ towards

the center of the circular path (Fig. Right hand side). The component Rcosθ

balances the weight *mg *and component Rsinθ provides the necessary centripetal force for the cyclist to move on the circular path. Now

So from above equation we conclude,

If the cyclist wands to turn the banking at higher speed speed he must bend more from the normal.

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Laws Of Rotational Motion (Part-2)

Laws Of Rotational Motion (Part-2)

Laws Of Rotational Motion (Part-2)

Laws Of Rotational Motion (Part-2)

Laws Of Rotational Motion (Part-2)