Mathematics THEOREMS OF PERPENDICULAR AND PARALLEL AXES, KINEMATICS AND DYNAMICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS
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CLASS 11 CHAPTER 7 - SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

THEOREMS OF PERPENDICULAR AND PARALLEL AXES, KINEMATICS AND DYNAMICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS

Topic covered

`star` THEOREMS OF PERPENDICULAR AND PARALLEL AXES
`star` KINEMATICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS
`star` DYNAMICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS

THEOREMS OF PERPENDICULAR AND PARALLEL AXES

`color{purple}bbul{✍️ "Theorem of perpendicular axes"}`

`=>` This theorem is applicable to bodies which are planar. In practice this means the theorem applies to flat bodies whose thickness is very small compared to their other dimensions (e.g. length, breadth or radius). Fig. 7.29 illustrates the theorem.

It states that the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.





`●` The figure shows a planar body. An axis perpendicular to the body through a point `O` is taken as the z-axis. Two mutually perpendicular axes lying in the plane of the body and concurrent with z-axis, i.e. passing through `O`, are taken as the x and y-axes. The theorem states that
`color{blue}{I_ z = I _x + I_y................... (7.36)}`

`=>` Let us look at the usefulness of the theorem through an example.

`color{purple}bbul{✍️ "Theorem of parallel axes"}`

`●` This theorem is applicable to a body of any shape. It allows to find the moment of inertia of a body about any axis, given the moment of inertia of the body about a parallel axis through the centre of mass of the body.
`●` We shall only state this theorem and not give its proof. We shall, however, apply it to a few simple situations which will be enough to convince us about the usefulness of the theorem. The theorem may be stated as follows:

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 its centre of mass and the product of its mass and the square of the distance between the two parallel axes.



`●` As shown in the Fig. 7.31, `z` and `z′ ` are two parallel axes separated by a distance `a`. The z-axis passes through the centre of mass `O` of the rigid body. Then according to the theorem of parallel axes

`color{blue}{I_(z')=I_z+Ma^2....................(7.37)}`

`=>` where `I_z` and `I_(z′)` are the moments of inertia of the body about the `z` and `z′` axes respectively, `M` is the total mass of the body and a is the perpendicular distance between the two parallel axes.

Q 3260456315

What is the moment of inertia of a disc about one of its diameters?

Solution:

We assume the moment of inertia of the disc about an axis perpendicular to it and through its centre to be known; it is `color{blue}{MR^2//2}`,
where `M` is the mass of the disc and `R` is its radius (Table 7.1)

The disc can be considered to be a planar body. Hence the theorem of perpendicular axes is applicable to it. As shown in Fig. 7.30, we
take three concurrent axes through the centre of the disc, `O` as the `x,y,z` axes; `x` and `y`-axes lie in the plane of the disc and z is perpendicular to it. By the theorem of perpendicular axes,

`color{blue}{i_z=I_x+I_y}`

Now, `x` and `y` axes are along two diameters of the disc, and by symmetry the moment of inertia of the disc is the same about any
diameter. Hence

`color{blue}{I_x=I_y}`

and `color{blue}{I_z=2I_x}`

But `color{blue}{I_z=MR^2//2}`

So finally, `color{blue}{I_x=I_z//2=MR^2//4}`

Thus the moment of inertia of a disc about any of its diameter is `color{blue}{MR^2//4} `.
Q 3280456317

What is the moment of inertia of a rod of mass `M`, length `l` about
an axis perpendicular to it through one end?

Solution:

For the rod of mass `M` and length `l`, `color{blue}{I = Ml^2//12}`. Using the parallel axes theorem,
`color{blue}{I′ = I + Ma^2}` with `color{blue}{a = l//2}` we get,

`color{blue}{I' =M(l^2)/(12)+M(l/2)^2=(Ml^2)/3}`

We can check this independently since `I` is half the moment of inertia of a rod of mass `2M`
and length `2l` about its midpoint,

`color{blue}{I'=2M * (4l^2)/12 xx 1/2=(Ml^2)/3}`
Q 3200456318

What is the moment of inertia of a ring about a tangent to the circle of the ring?

Solution:

The tangent to the ring in the plane of the ring is parallel to one of the diameters of the ring.

The distance between these two parallel axes is `R`, the radius of the ring. Using the parallel axes
theorem,

`color{blue}{I_("tangent")=I_("dia")+MR^2=(MR^2)/2 +MR^2=3/2 MR^2}`

KINEMATICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS

`●` Let angular velocity `ω` plays the same role in rotation as the linear velocity `v` in translation. We wish to take this analogy further. In doing so we shall restrict the discussion only to rotation about fixed axis.
`●` This case of motion involves only one degree of freedom, i.e., needs only one independent variable to describe the motion. This in translation corresponds to linear motion. This section is limited only to kinematics. We shall turn to dynamics in later sections.

`●` We recall that for specifying the angular displacement of the rotating body we take any particle like P (Fig.7.33) of the body. Its angular displacement `θ` in the plane it moves is the angular displacement of the whole body; `θ` is measured from a fixed direction in the plane of motion of `P`, which we take to be the `x′` - axis, chosen parallel to the `x`-axis. Note, as shown, the axis of rotation is the `z` – axis and the plane of the motion of the particle is the `x - y` plane. Fig. 7.33 also shows `θ_o`, the angular displacement at `t = 0`.

`●` We also recall that the angular velocity is the time rate of change of angular displacement, `ω = dθ//dt`. Note since the axis of rotation is fixed, there is no need to treat angular velocity as a vector. Further, the angular acceleration, `alpha=(d omega//dt)`

`●` The kinematical quantities in rotational motion, angular displacement `(θ)`, angular velocity `(ω)` and angular acceleration `(α)` respectively correspond to kinematic quantities in linear motion, displacement `(x)`, velocity `(v)` and acceleration `(a)`. We know the kinematical equations of linear motion with uniform (i.e. constant) acceleration:

`color{blue}{v=v_0+at.......................(a)}`


`color{blue}{x=x_0+v_0t+1/2 at^2...............(b)}`

`color{blue}{v^2=v_0^2+2ax...................(c)}`

`=>` where `x_0 =` initial displacement and `v_0=` initial velocity. The word ‘initial’ refers to values of the quantities at `t = 0`

`●` The corresponding kinematic equations for rotational motion with uniform angular acceleration are:

`color{blue}{omega=omega_0 +alpha t..................(7.38)}`

`color{blue}{ theta=theta_0 +omega_0 t +1/2 alpha t^2.............(7.39)}`

and `color{blue}{omega^2=omega_0^2 + 2 alpha(theta-theta_0).............(7.40)}`

where `θ_0=` initial angular displacement of the rotating body, and `ω_0 =` initial angular velocity of the body.


Q 3220556411

Obtain Eq. (7.38) `color{blue}{omega=omega_0 +alpha_t}` from first principles.

Solution:

The angular acceleration is uniform,

hence

`color{blue}{(d omega)/(dt) = alpha=" constant....................(1)"}`

Integrating this equation,

`color{blue}{omega= int alpha dt+c}`

`=alpha t +c` (as `α` is constant)

At `t = 0, ω = ω_0` (given)

From (i) we get at `t = 0, ω = c = ω_0`

Thus, `ω = αt + ω_0` as required.

With the definition of `color{blue}{ω = dθ//dt}` we may integrate Eq. (7.38) to get Eq. (7.39). This derivation and the derivation of Eq. (7.40) is
left as an exercise.
Q 3260178915

The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm in 16 seconds. (i) What is its angular acceleration, assuming the acceleration to be uniform? (ii) How many revolutions does the engine make during this time?
Class 11 Chapter 7 Example 14
Solution:

We shall use `omega = omega_0 + a t `

`omega_0 =` Initial angular speed in rad/s

` = 2 pi xx` Angular speed in rev/s

` = (2 pi xx "Angular speed in rev"//min)/(60 s//min)`

` = (2 pi xx 1200)/(60)` rad/s `= 40 pi` rad/ s

Similarly, `omega =` Final angular speed

`(2 pi xx 3120)/(60) r = 104 pi` rad /s

: . Angular acceleration

`alpha = (omega - omega_0)/t = (104 pi - 40 pi)/(16) = 4 pi rad//s^2`

The angular acceleration of the motor `= 4 pi rad//s^2`

(ii) The angular displacement in time t is given by

`theta = omega_0 t + 1/2 alpha t^2`

` = (40 pi xx 16 + 1/2 xx 4 pi + 16^2)`

` = (640 pi + 512 pi) = 1152 pi rad`

Number of revolutions `= (1152 pi)/(2 pi) = 576`

DYNAMICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS

`●` Table 7.2 lists quantities associated with linear motion and their analogues in rotational motion. Also, we know that in rotational motion moment of inertia and torque play the same role as mass and force respectively in linear motion. Given this we should be able to guess what the other analogues indicated in the table are.
`●` For example, we know that in linear motion, work done is given by `F dx`, in rotational motion about a fixed axis it should be `τ` dθ , since we already know the correspondence `dx → dθ` and `F →τ` . It is, however, necessary that these correspondences are established on sound dynamical considerations. This is what we now turn to.

`●` Before we begin, we note a simplification that arises in the case of rotational motion about a fixed axis. Since the axis is fixed, only those components of torques, which are along the direction of the fixed axis need to be considered in our discussion.
`●` Only these components can cause the body to rotate about the axis. A component of the torque perpendicular to the axis of rotation will tend to turn the axis from its position.
`●` We specifically assume that there will arise necessary forces of constraint to cancel the effect of the perpendicular components of the (external) torques, so that the fixed position of the axis will be maintained. The perpendicular components of the torques, therefore need not be taken into account. This means that for our calculation of torques on a rigid body:

(1) We need to consider only those forces that lie in planes perpendicular to the axis. Forces which are parallel to the axis will give torques perpendicular to the axis and need not be taken into account.

(2) We need to consider only those components of the position vectors which are perpendicular to the axis. Components of position vectors along the axis will result in torques perpendicular to the axis and need not be taken into account.
 
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