Overview: Rotational Dynamics

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

F = mω²r = \[\frac {mv^2}{r}\] = mrw

a_r = ω²r

• 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.

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.

A uniform ring: I = MR²

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

I = MK²

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.

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

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

Magnitude Formula: L = r p sin θ

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.

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