Mathematics Centre of gravity AND MOMENT OF INERTIA

Topic covered

`star` Centre of gravity

Centre of gravity

`●` Figure 7.24 illustrates a similar experiment that you can easily perform. Take an irregularshaped cardboard and a narrow tipped object like a pencil. You can locate by trial and error a point `G` on the cardboard where it can be balanced on the tip of the pencil. (The cardboard remains horizontal in this position.)

`●` This point of balance is the centre of gravity (`CG`) of the cardboard.
`●` The tip of the pencil provides a vertically upward force due to which the cardboard is in mechanical equilibrium.

`●` As shown in the Fig. 7.24, the reaction of the tip is equal and opposite to `Mg`, the total weight of (i.e., the force of gravity on) the cardboard and hence the cardboard is in translational equilibrium. It is also in rotational equilibrium; if it were not so, due to the unbalanced torque it would tilt and fall.
`●` There are torques on the card board due to the forces of gravity like `m_1g, m_2g ….` etc, acting on the individual particles that make up the cardboard.

`●` The `CG` of the cardboard is so located that the total torque on it due to the forces `m_1g, m_2g ….` etc. is zero.
`●` If `r_i` is the position vector of the ith particle of an extended body with respect to its CG, then the torque about the CG, due to the force of gravity on the particle is `τ_ i = r_i × m_i g`. The total gravitational torque about the `CG` is zero, i.e.

`color{blue}{tau_g = sum tau_i= sum r_i xx m_i g =0..................(7.33)}`

`\color{red} ✍️` We may therefore, define the CG of a body as that point where the total gravitational torque on the body is zero.

`●` We notice that in Eq. (7.33), `g` is the same for all particles, and hence it comes out of the summation. This gives, since `g` is nonzero, `sum m_i r_i =0`.

`bbul{"Remember"}` that the position vectors (ri) are taken with respect to the CG. Now, in accordance with the reasoning given below Eq. (7.4a) in Sec. 7.2, if the sum is zero, the origin must be the centre of mass of the body. Thus, the centre of gravity of the body coincides with the centre of mass.

`●` We note that this is true because the body being small, `g` does not vary from one point of the body to the other. If the body is so extended that g varies from part to part of the body, then the centre of gravity and centre of mass will not coincide.
`●` Basically, the two are different concepts. The centre of mass has nothing to do with gravity. It depends only on the distribution of mass of the body.

`●` In Sec. 7.2 we found out the position of the centre of mass of several regular, homogeneous objects. Obviously the method used there gives us also the centre of gravity of these bodies, if they are small enough.

Figure 7.25 illustrates another way of determining the `CG` of an regular shaped body like a cardboard. If you suspend the body from some point like `A`, the vertical line through A passes through the `CG`.
`●` We mark the vertical `A A_1`. We then suspend the body through other points like `B` and `C`. The intersection of the verticals gives the CG. Explain why the method works. Since the body is small enough, the method allows us to determine also its centre of mass.
Q 3230178912

A metal bar 70 cm long and 4.00 kg in mass supported on two knife-edges placed 10 cm from each end. A 6.00 kg load is suspended at 30 cm from one end. Find the reactions at the knifeedges. (Assume the bar to be of uniform cross section and homogeneous.)
Class 11 Chapter 7 Example 8

Figure 7.26 shows the rod AB, the positions of the knife edges `K_1` and `K_2` , the centre of gravity of the rod at G and the suspended load at P.

Note the weight of the rod W acts at its centre of gravity G. The rod is uniform in cross section and homogeneous; hence G is at the centre of the rod; AB = 70 cm. AG = 35 cm, AP = 30 cm, PG = 5 cm, `AK_1= BK_2 = 10` cm and `K_1G = K_2G = 25` cm.

Also, W= weight of the rod = 4.00 kg and `W_1 =` suspended load = 6.00 kg; `R_1` and `R_2` are the normal reactions of the support at the knife edges.

For translational equilibrium of the rod,
`color{orange} {R_1 + R_2 –W_1 –W = 0}` ........... (i)

Note `W_1` and W act vertically down and `R_1`

and `R_2` act vertically up.

For considering rotational equilibrium, we take moments of the forces. A convenient point to take moments about is G. The moments of `R_2` and `W_1` are anticlockwise (+ve), whereas the moment of `R_1` is clockwise (-ve).

For rotational equilibrium,

`color{green} {–R_1 (K_1G) + W_1 (PG) + R_2 (K_2G) = 0}` ......... (ii)

It is given that W = 4.00g N and W1 = 6.00g

N, where g = acceleration due to gravity. We take g = 9.8 m/s2. With numerical values inserted, from (i)

`R_1 + R_2 – 4.00g – 6.00g = 0`

or `R_1 + R_2 = 10.00g N` .................. (iii)

= 98.00 N

From (ii), `– 0.25 R_1 + 0.05 W_1 + 0.25 R_2 = 0`

or `R_1 – R_2 = 1.2g N = 11.76 N` ............... (iv)

From (iii) and (iv), `R_1 = 54.88 N`,

`R_2 = 43.12 N`

Thus the reactions of the support are about 55 N at `K_1` and 43 N at `K_2`.
Q 3250178914

A 3m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor 1 m from the wall as shown in Fig.7.27. Find the reaction forces of the wall and the floor.
Class 11 Chapter 7 Example 9

The ladder AB is 3 m long, its foot A is at distance AC = 1 m from the wall. From Pythagoras theorem, `BC = 2sqrt2` m .

The forces on the ladder are its weight W acting at its centre of gravity D, reaction forces `F_1` and `F_2` of the wall and the floor respectively. Force `F_1` is perpendicular to the wall, since the wall is frictionless. Force `F_2` is resolved into two components, the normal reaction N and the force of friction F. Note that F prevents the ladder from sliding away from the wall and is therefore directed toward the wall.

For translational equilibrium, taking the forces in the vertical direction,

`N – W = 0` ........ (i)

Taking the forces in the horizontal direction,

`F – F_1 = 0` ................ (ii)

For rotational equilibrium, taking the moments of the forces about A,

`2sqrt2F_1 - (1/2) W = 0` ...................(iii)

Now W = 20 `g = 20 × 9.8 N = 196.0 N`

From (i) N = 196.0

From (iii) `F_1 =W//4sqrt2 = 196.0//4sqrt2 =34.6 N`

From (ii) `F = F_1 = 34.6 N`

`F_2 = sqrt(F^2 +N^2) = 199.0 N`

The force `F_2` makes an angle a with the horizontal,

`tan alpha = N//F = 4 sqrt2, alpha = tan^-1(4sqrt2) ~~ 80^o`


`●` To keep the discussion simple, we shall consider rotation about a fixed axis only. Let us try to get an expression for the kinetic energy of a rotating body.
`●` We know that for a body rotating about a fixed axis, each particle of the body moves in a circle with linear velocity given by Eq. (7.19). (Refer to Fig. 7.16). For a particle at a distance from the axis, the linear velocity is `υ_i = rω` . The kinetic energy of motion of this particle is

`color{orange}{k_i = 1/2 m_i v_i^2= 1/2 m_i r_i^2 ω^2}`

`=>` where `m_i` is the mass of the particle. The total kinetic energy `K` of the body is then given by the sum of the kinetic energies of individual particles,

`K= sum_(i=1)^n k_i= 1/2 sum_(i=1)^n (m_i r_i^2 omega^2)`

`●` Here `n` is the number of particles in the body. Note `ω` is the same for all particles. Hence, taking `ω` out of the sum,

`K=1/2 omega^2(sum_(i=1)^n m_i r_i^2)`

`●` We define a new parameter characterising the rigid body, called the moment of inertia `I` , given by

`color{blue}{I=sum_(i=1)^n m_i r_i^2...................(7.34)}`

With this definition,

`color{blue}{K= 1/2 I omega^2....................(7.35)}`

`●` Note that the parameter `I` is independent of the magnitude of the angular velocity. It is a characteristic of the rigid body and the axis about which it rotates.
Compare Eq. (7.35) for the kinetic energy of a rotating body with the expression for the kinetic energy of a body in linear (translational) motion,

`K= 1/2 mv^2`

`=>` Here `m` is the mass of the body and v is its velocity. We have already noted the analogy between angular velocity `ω` (in respect of rotational motion about a fixed axis) and linear velocity `v` (in respect of linear motion).
`=>` It is then evident that the parameter, moment of inertia I, is the desired rotational analogue of mass. In rotation (about a fixed axis), the moment of inertia plays a similar role as mass does in linear motion.

We now apply the definition Eq. (7.34), to calculate the moment of inertia in two simple cases.

(a) Consider a thin ring of radius `R` and mass `M`, rotating in its own plane around its centre with angular velocity `ω`. Each mass element of the ring is at a distance `R` from the axis, and moves with a speed `Rω`. The kinetic energy is therefore,

`K=1/2 Mv^2=1/2 MR^2 omega^2`

Comparing with Eq. (7.35) we get `I = MR^2` for the ring.

(b) Next, take a rigid massless rod of length `l` with a pair of small masses, rotating about an axis through the centre of mass perpendicular to the rod (Fig. 7.28). Each mass M/2 is at a distance l/2 from the axis. The moment of inertia of the masses is therefore given by

`(M//2) (l//2)^2 + (M//2)(l//2)^2`

`=>` Thus, for the pair of masses, rotating about the axis through the centre of mass perpendicular to the rod

`I = Ml^2 // 4`

Table 7.1 gives the moment of inertia of various familiar regular shaped solids about specific axes.

`●` As the mass of a body resists a change in its state of linear motion, it is a measure of its inertia in linear motion.
`●` Similarly, as the moment of inertia about a given axis of rotation resists a change in its rotational motion, it can be regarded as a measure of rotational inertia of the body; it is a measure of the way in which different parts of the body are distributed at different distances from the axis.
`●` Unlike the mass of a body, the moment of inertia is not a fixed quantity but depends on the orientation and position of the axis of rotation with respect to the body as a whole.
`●` As a measure of the way in which the mass of a rotating rigid body is distributed with respect to the axis of rotation, we can define a new parameter, the radius of gyration. It is related to the moment of inertia and the total mass of the body.

`\color{green} ✍️` Notice from the Table 7.1 that in all cases, we can write `I = Mk^2`, where `k` has the dimension of length. For a rod, about the perpendicular axis at its midpoint, `k^2=L^2//12` i.e. `k^2=L//sqrt 12`

`●` Similarly, `k = R//2` for the circular disc about its diameter. The length `k` is a geometric property of the body and axis of rotation. It is called the radius of gyration. The radius of gyration of a body about an axis may be defined as the distance from the axis of a mass point whose mass is equal to the mass of the whole body and whose moment of inertia is equal to the moment of inertia of the body about the axis.

`\color{red} ✍️` Thus, the moment of inertia of a rigid body depends on the mass of the body, its shape and size; distribution of mass about the axis of rotation, and the position and orientation of the axis of rotation.

From the definition, Eq. (7.34), we can infer that the dimensions of moments of inertia we `ML^2` and its SI units are `kg m^2`.

`●` The property of this extremely important quantity `I` as a measure of rotational inertia of the body has been put to a great practical use. The machines, such as steam engine and the automobile engine, etc., that produce rotational motion have a disc with a large moment of inertia, called a flywheel.
`●` Because of its large moment of inertia, the flywheel resists the sudden increase or decrease of the speed of the vehicle. It allows a gradual change in the speed and prevents jerky motions, thereby ensuring a smooth ride for the passengers on the vehicle.