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

The circular motion in which the speed of a particle is constant but its direction changes continuously, and acceleration is always directed towards the centre, is called uniform circular motion.

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 non-real (fictitious) force directed along the radius away from the centre of a circle (opposite to centripetal acceleration) is called centrifugal force.

The force directed along the radius towards the centre of a circle, which is necessary to keep the object moving in a circle, is called centripetal force.

When a particle moves in two dimensions or in a plane such that its distance from a fixed (or moving) point remains constant, then its motion is called circular motion.

The vertical cylindrical wall along which motorcycles or cars move in a long horizontal circle, using normal reaction force from the wall as centripetal force, is called the well of death.

The angle by which a road should be banked to allow safe circular motion of a vehicle is called the banking angle.

The road that is banked at a certain angle (θ) to avoid the skidding of vehicles travelling on a circular path is called a banked road.

A mass suspended by a string that rotates in a horizontal circular path, forming a cone shape, is called a conical pendulum.

A special type of circular motion in which the body travels in a vertical plane is called vertical circular motion.

The sum of the product of the mass of each particle and the square of its perpendicular distance from the axis of rotation is called the moment of inertia of a rigid body about an axis of rotation.

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.

The distance between the axis of rotation and a point at which the entire mass of the body can be supposed to be concentrated so as to possess the same moment of inertia as that of the body about that axis is called the radius of gyration of a rigid body about a given axis of rotation.

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`

The quantity in rotational mechanics that is analogous to linear momentum is called angular momentum.

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 turning effect of a force about the axis of rotation is called moment of force or torque due to that force.

When a body performs translational motion as well as rotational motion simultaneously, this type of motion is called rolling motion.

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

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.

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.

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

\[\vec{F}=-\frac{mv^2}{r}\hat{r}_0\]

Directed towards the centre (negative sign indicates inward direction).

\[\vec{F}=+\frac{mv^2}{r}\hat{r}_0\]

Directed away from the centre (positive sign indicates outward direction).

v_s = \[\sqrt {μrg}\]

where μ = coefficient of friction, r = radius, g = acceleration due to gravity.

v = \[\sqrt {rg tan θ}\]

\[\theta=\tan^{-1}\left(\frac{v^2}{rg}\right)\]

\[v_{\min}=\sqrt{rg\left(\frac{\tan\theta-\mu_s}{1+\mu_s\tan\theta}\right)}\]

\[v_{\max}=\sqrt{rg\left(\frac{\tan\theta+\mu_s}{1-\mu_s\tan\theta}\right)}\]

\[v_{\min}=\sqrt{\frac{rg}{\mu_s}}\]

T_B​ = 6mg + T_A​

T_D ​= 3mg + T_A​

T_C ​= T_D​

Total KE = Translational KE + Rotational KE

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

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

Statement: The moment of inertia (I_z​) of a laminar object about an axis (z) perpendicular to its plane is equal to the sum of its moment of inertias about two mutually perpendicular axes (x and y) in its plane, all three axes being concurrent.

Statement: The moment of inertia (I_o​) of an object about any axis is equal to the sum of its moment of inertia (I_c) about an axis parallel to the given axis and passing through the centre of mass, and the product of the mass of the object and the square of the distance between the two axes.

where M = mass of the object, h = distance between the two parallel axes.

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.

Angular momentum of an isolated system is conserved in the absence of an external unbalanced torque.

L = constant(when τ_ext​ = 0)

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 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​

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.

• Tension — Minimum at topmost point A, intermediate at C & D, and maximum at bottommost point B.
• Velocity — Minimum at topmost point A (rg) and maximum at bottommost point B (5rg).
• Motion — Body accelerates from A to B (left/C side) and decelerates from B to A (right/D side).

• Thin ring / Hollow cylinder (Central axis) → I = MR²
• Thin ring (Diameter) → I = \[\frac {1}{2}\]M R²
• Annular ring / Thick hollow cylinder (Central) → I = \[\frac{1}{2}M (R_1^2+R_2^2)\]
• Uniform disc / Solid cylinder (Central) → I = \[\frac {1}{2}\]M R²
• Uniform disc (Diameter) → I = \[\frac {1}{4}\]M R²
• Solid sphere (Central) → I = \[\frac {2}{5}\]M R²
• Thin walled hollow sphere (Central) → I = \[\frac {2}{3}\]M R²
• Thin rod (Centre, ⊥ to length) → I = \[\frac {1}{12}\]M L²
• Thin rod (One end, ⊥ to length) → I = \[\frac {1}{3}\]M L²
• Solid cone (Central) → I = \[\frac {3}{10}\]M R², Hollow cone (Central) → I = \[\frac {1}{2}\]M R²

• 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.