- A person on the ground sees the ball move in a circle due to centripetal force (tension in the string).
- A person on the rotating platform sees the ball as stationary only if they assume an opposite force to balance the tension.
- When the string breaks, the ground observer sees the ball go straight, while the rotating observer sees it move outward, due to the assumed centrifugal force.
Definitions [22]
Definition: Centrifugal Force
The non-real (fictitious) force directed along the radius away from the centre of a circle (opposite to centripetal acceleration) is called centrifugal force.
Definition: Circular Motion
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.
Definition: Uniform Circular Motion
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.
Definition: Centripetal 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.
Define centripetal force.
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
Definition: Non-Uniform Circular Motion
The circular motion in which the speed of a particle changes at different points along the path and tangential acceleration is present due to change in speed, is called non-uniform circular motion.
Definition: Centrifugal Force
A force assumed (by an observer moving with the body) to be acting on the body in a direction away from the centre of a circular path is called centrifugal force.
Definition: Fictitious Force
A force which really does not exist, but is considered to describe (or understand) a certain motion, is called a fictitious force (or virtual force).
Definition: Well of Death
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.
Definition: Banking Angle
The angle by which a road should be banked to allow safe circular motion of a vehicle is called the banking angle.
Definition: Banked Road
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.
Definition: Conical Pendulum
A mass suspended by a string that rotates in a horizontal circular path, forming a cone shape, is called a conical pendulum.
Definition: Conical Pendulum
A simple pendulum which is given such a motion that the bob describes a horizontal circle and the string makes a constant angle with the vertical, such that the string describes a cone, is called a conical pendulum.
Definition: Vertical Circular Motion
A special type of circular motion in which the body travels in a vertical plane is called vertical circular motion.
Definition: Moment of Inertia
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.
Definition: Radius of Gyration
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.
K = \[\sqrt {\frac {I}{M}\], I = MK2
Define radius of gyration.
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.
Define angular momentum.
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`
Define moment of inertia.
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`
Definition: Angular Momentum
The quantity in rotational mechanics that is analogous to linear momentum is called angular momentum.
Definition: Torque
The turning effect of a force about the axis of rotation is called moment of force or torque due to that force.
Definition: Rolling Motion
When a body performs translational motion as well as rotational motion simultaneously, this type of motion is called rolling motion.
OR
The combination of rotational and translational motion of a round-shaped body when the body is in contact with a surface, such that every particle of the body has two velocities (one due to rotation and one due to translation) whose vector sum gives the resultant velocity, is called Rolling Motion.
Formulae [14]
Formula: Centrifugal Force
\[\vec{F}=+\frac{mv^2}{r}\hat{r}_0\]
Directed away from the centre (positive sign indicates outward direction).
Formula: Centripetal Force
\[\vec{F}=-\frac{mv^2}{r}\hat{r}_0\]
Directed towards the centre (negative sign indicates inward direction).
Formula: Banked Road - Most Safe Speed
v = \[\sqrt {rg tan θ}\]
Formula: Banking Angle
\[\theta=\tan^{-1}\left(\frac{v^2}{rg}\right)\]
Formula: Banked Road - Minimum Speed
\[v_{\min}=\sqrt{rg\left(\frac{\tan\theta-\mu_s}{1+\mu_s\tan\theta}\right)}\]
Formula: Banked Road - Maximum Speed
\[v_{\max}=\sqrt{rg\left(\frac{\tan\theta+\mu_s}{1-\mu_s\tan\theta}\right)}\]
Formula: Well of Death - Minimum Speed
\[v_{\min}=\sqrt{\frac{rg}{\mu_s}}\]
Formula: Unbanked Road - Maximum Speed
vs = \[\sqrt {μrg}\]
where μ = coefficient of friction, r = radius, g = acceleration due to gravity.
Formula: Time Period of Conical Pendulum
T = 2π\[\sqrt{\frac{l\cos\theta}{g}}\]
Forces acting on the bob:
- T cos θ — balances the weight mg (vertical component)
- T sin θ — provides the centripetal force (horizontal component)
Formula: Tension Relations
TB = 6mg + TA
TD = 3mg + TA
TC = TD
Formula: Minimum Velocities
| Position | Minimum Velocity |
|---|---|
| Topmost Point A | vmin = \[\sqrt {rg}\] |
| Middle Points C & D | vmin = \[\sqrt {3rg}\] |
| Bottommost Point B | vmin = \[\sqrt {5rg}\] |
Formula: Equation at Each Position
| Position | Equation |
|---|---|
| Topmost Point A | mg + TA = \[\frac {mv^2A}{r}\] |
| Bottommost Point B | TB − mg = \[\frac {mv^2B}{r}\] |
| Middle Points C & D | TC − TA = TD − TA = 3mg |
Formula: Total Kinetic Energy of Rolling
Total KE = Translational KE + Rotational KE
Formula: Velocity at Point
- Velocity at point A (top): vA = v + Rω = 2v (or \[\sqrt 2\]Rω)
- Velocity at point B (side): vB = \[\sqrt {v^2+(Rω)^2}\] = \[\sqrt 2\]Rω
- Velocity at point C (bottom): vC = v − Rω = 0
- Velocity at point D (side): vD = \[\sqrt {v^2+(Rω)^2}\] = \[\sqrt 2\] Rω
Theorems and Laws [6]
State and prove the theorem of the parallel axis about the moment of inertia.
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 ICM 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 = ICM + Mh2
This proves the theorem of the parallel axis.
Theorem: Perpendicular Axis Theorem
Statement: The moment of inertia (Iz) 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.
Iz = Ix + Iy
Theorem: Parallel Axis Theorem
Statement: The moment of inertia (Io) of an object about any axis is equal to the sum of its moment of inertia (Ic) 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.
Io = Ic + Mh2
where M = mass of the object, h = distance between the two parallel axes.
Law: Conservation of Angular Momentum
Angular momentum of an isolated system is conserved in the absence of an external unbalanced torque.
L = constant(when τext = 0)
State and prove: Law of conservation of angular momentum.
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.
Law: Conservation of Angular Momentum
Statement: The total angular momentum of an isolated system remains constant (conserved) if no resultant external torque acts on it.
Mathematical Form:
If \[\vec τ_{ext}\] = 0, then:
\[\frac {d\vec L}{dt}\] = 0 \[\vec L\] = constant
I1ω1 = I2ω2
Key Points:
- Angular momentum \[\vec L\] = I\[\vec ω\]
- When net external torque is zero, angular momentum remains constant.
- Angular impulse = net torque × time = change in angular momentum: τ ⋅ Δt = ΔL
Key Points
Key Points: Centrifugal Force
Key Points: Conical Pendulum
- Structure — A conical pendulum consists of a weight (bob) fixed on a string suspended from a pivot, where the bob revolves in a horizontal circle and the string describes a cone whose vertex is the fixed point.
- Time Period depends on — Length of pendulum (l), acceleration due to gravity (g), and angle of inclination (θ); given by T = 2π\[\sqrt{\frac{l\cos\theta}{g}}\].
- Force Analysis — The vertical component of tension (T cos θ) balances gravity (mg), while the horizontal component (T sin θ) acts as the centripetal force to maintain circular motion.
Key Points: Vertical Circular Motion
- 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).
Key Points: M.I. of Symmetrical Objects
- 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
Concepts [22]
- Rotational Dynamics
- Circular Motion and Its Characteristics
- Kinematics of Rotational Motion About a Fixed Axis
- Non-uniform circular motion
- Dynamic of Circular Motion
- Centrifugal Forces
- Applications of Uniform Circular Motion
- Vehicle Along a Horizontal Circular Track
- Well (or Woll) of Death
- Vehicle on o Banked Rood
- Conical Pendulum
- Vertical Circular Motion
- Point Moss Undergoing Vertical Circular Motion Under Gravity
- Sphere of Death
- Moment of Inertia as an Analogous Quantity for Mass
- Moment of Inertia of a Uniform Disc
- Radius of Gyration
- Theorems of Perpendicular and Parallel Axes
- Angular Momentum or Moment of Linear Momentum
- Expression for Torque in Terms of Moment of Inertia
- Conservation of Angular Momentum
- Rolling Motion
