Revision: Rotational Dynamics Physics HSC Science (General) 12th Standard Board Exam Maharashtra State Board

The force acting on a particle performing uniform circular motion along the radius and directed towards the centre of the circle is called the centripetal force.

The mathematical form of centripetal force is:

F = `mv^2/r`

where:

F = centripetal force,

m = mass of the object,

v = speed or velocity, and

r = radius

The radius of gyration of a body is defined as the distance between the axis of rotation and a point at which the whole mass of the body is supposed to be concentrated, so as to possess the same moment of inertia as that of the body.

Moment of Inertia (I) is the measure of an object's resistance to changes in its rotational motion about a given axis. It is defined as the sum of the products of the masses of the individual particles of the body and the square of their perpendicular distances from the axis of rotation. Mathematically, it is expressed as:

`I = summ_ir2/i`

The rotational mechanics equivalent of linear momentum is angular momentum or moment of linear momentum. It is analogous to the torque being the moment of force.

i.e `vecL = vecr xxvecP`

Moment of inertia of a body about a given axis is defined as the sum of the products of each mass element and the square of its perpendicular distance from the axis of rotation.

During circular motion, if the speed of the particle remains constant, it is called Uniform Circular Motion (UCM).

The quantity in rotational mechanics, analogous to linear momentum, is angular momentum or moment of linear momentum. 

Rolling motion is the motion in which a body simultaneously undergoes translational motion of its centre of mass and rotational motion about its own axis.

F = mω²r = \[\frac {mv^2}{r}\] = mrw

a_r = ω²r

Vector Formula: \[\vec L\] = \[\vec r\] × \[\vec p\]

Magnitude Formula: L = r p sin θ

A uniform ring: I = MR²

A uniform disc: I = \[\frac {1}{2}\]MR²

I = MK²

\[{E=\frac{1}{2}Mv^2\left(1+\frac{K^2}{R^2}\right)}\]

A body's moment of inertia along an axis is equal to the product of two things: Its moment of inertia about a parallel axis through its centre of mass and the product of the body's mass and the square of the distance between the two axes. This is known as the parallel axis theorem.

Proof: Let I_CM represent a body of mass M moment of inertia (MI) about an axis passing through its centre of mass C and let I stand for that body's MI about a parallel axis passing through any point O. Let h represent the separation of the two axis.

Think about the body's minuscule mass element dm at point P. It is perpendicular to the rotation axis through point C and to the parallel axis through point O, with a corresponding perpendicular distance of OP. CP² dm is the MI of the element about the axis through C. As a result, `I_(CM) = int CP^2  dm` is the body's MI about the axis through the CM. In a similar vein, `I = int OP^2  dm` is the body's MI about the parallel axis through O.

Draw PQ perpendicular to OC produced, as shown in the figure. Then, from the figure,

`I = int OP^2  dm`

= `int (OQ^2 + PQ^2) dm`

= `int [(OC + CQ)^2 + PQ^2] dm`

= `int (OC^2 + 2OC.CQ + CQ^2 + PQ^2) dm`

= `int (OC^2 + 2OC.CQ + CP^2)dm`   ...(∵ CQ² + PQ² = CP²)

= `int OC^2  dm + int 2OC.CQ  dm + int CP^2  dm`

= `OC^2 int dm + 2OC int CQ  dm + int CP^2  dm`

Since OC = h is constant and `int dm = M` is the mass of the body,

`I = Mh^2 + 2h int CQ  dm + I_(CM)`

The integral `int CQ  dm` now yields mass M times a coordinate of the CM with respect to the origin C, based on the concept of the centre of mass. This position and the integral are both zero because C is the CM in and of itself.

∴ I = I_CM + Mh²

This proves the theorem of the parallel axis.

Statement:

The angular momentum of a body remains constant if the resultant external torque acting on the body is zero.

I₁ω₁ = I₂ω₂ (when τ = 0)

Here I is the moment of inertia and ω is angular velocity.

Proof:

Consider a particle of mass m, rotating about an axis with torque ‘τ’.

Let `vecp` be the linear momentum of the particle and `vecr` be its position vector.

∴ Angular momentum, `vecL = vecr xx vecp`  .....(1)

Differentiating equation (1) with respect to time t, we get,

`(dvecL)/(dt) = d/dt(vecr xx vecp) = vecr(dvecp)/(dt) + vecp(dvecr)/(dt)`

We know that, `(dvecp)/(dt) = vecF, (dvecr)/(dt) = vec"v", vecp = mvec"v"`

∴ `(dvecL)/(dt) = vecr xx vecF + m(vec"v" xx vec"v")`

∴ `(dvecL)/(dt) = vecr xx vecF`  ....`(∵ vec"v" xx vec"v" = 0)`

∴ `(dvecL)/(dt) = vectau` ......`(∵ vecr xx vecF = vectau)`

Now, If the `vectau = 0`, then

`(dvecL)/(dt) = 0`

∴ `vecL` is constant. Hence angular momentum remains conserved.

Example:

An athlete diving off a high springboard can bring his legs and hands close to the body and performs Somersault about a horizontal axis passing through his body in the air before reaching the water below it. During the fall, his angular momentum remains constant.

Statement

The moment of inertia of a plane laminar body about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two mutually perpendicular axes in its plane, provided all three axes are concurrent.

I_z = I_x + I_y​

Proof

Consider a plane lamina lying in the xy-plane.

Let:

• x-axis and y-axis lie in the plane of the lamina.

• The z-axis is perpendicular to the plane.

• All three axes intersect at the same point.

Take a small mass element dmdmdm at point P(x,y).

Distance of dmdmdm from:

• x-axis = y
• y-axis = x
• z-axis = \[\sqrt{x^2+y^2}\]​

Now,

I_x = ∫y² dm

I_y = ∫x² dm

I_z = ∫(x² + y²) dm

I_z = ∫x²dm + ∫y²dm​

I_z = I_y + I_x​

Hence,

I_z = I_x + I_y​

Conclusion

I_z = I_x + I_y​​

Thus, the moment of inertia about the perpendicular axis equals the sum of moments of inertia about the two in-plane perpendicular axes.

Statement

The moment of inertia of a body about any axis is equal to the moment of inertia about a parallel axis through its centre of mass plus Mh², where h is the distance between the two axes.

I = I_C + Mh²

Proof

Consider a body of mass M.
Let I_C​ be the moment of inertia about an axis through the centre of mass.
Let another axis be parallel to it at a distance h.

Take a small mass element dm.

I_O = ∫(r + h)²dm

Expanding:

I_O = ∫(r² + 2rh + h²)dm

I_O = ∫r²dm + 2h ∫rdm + h² ∫dm

Now,

∫ r²dm = I_C​

∫ rdm = 0(by definition of centre of mass)

∫ dm = M

Therefore,

I_O = I_C + Mh²

Conclusion

I = I_C + Mh²​

• On a horizontal road, static friction provides centripetal force, limiting maximum safe speed.
• In a “well of death,” a normal reaction provides centripetal force, and friction balances the weight.
• Banking roads reduce frictional losses and improve safety during turns at a given speed.
• On a banked road, the direction of friction changes depending on whether the speed is below or above the safe speed.
• In a conical pendulum, tension has two roles — the vertical component balances the weight, and the horizontal component provides the centripetal force.

• In vertical circular motion under gravity, speed continuously changes due to the conversion between kinetic and potential energy.
• At the topmost point, tension is minimum; at the bottommost point, tension is maximum.
• A string requires a minimum speed at the top to remain taut, but a rigid rod does not.
• At the top of a convex bridge, normal reaction decreases with speed and becomes zero when contact is just lost.
• Motion is non-uniform, so linear and angular accelerations are not constant and standard kinematic equations do not apply.