Rolling Motion | Physics for JEE Main & Advanced PDF Download

You must have observed the motion of wheels used in transportation at one or the other time in your life. In this document, we will be studying the motion of bodies like a wheel, which undergoes simultaneous rotational and translational motion, in detail.On looking closely, it can be seen that the motion of a wheel is a combination of straight line motion (translational motion) and rotational motion as shown in the animation below:

Motion of a Wheel

What is Rollin Motion?

Rolling motion is defined as the combined translational and rotational motion of an object. In rolling motion, an object moves along a surface while also rotating about its own axis. This type of motion is commonly observed in objects such as wheels, cylinders, or balls rolling on a surface.

Concept of Rolling Motion

• Imagine a car driving on a straight road. If we watch one of its wheels while standing still on the road, it seems like a mix of spinning and moving forward. 
• Now, let's pretend the car has its special frame attached, moving at the same speed as the car itself. In this attached frame, the wheel only seems to spin, and the whole car seems to just move without any spinning.
• Think about a fan inside the car. If the car is parked, the fan's motion is only spinning. If the car is moving, the fan's motion is a mix of spinning and moving in the car. 
• Now, if we're sitting inside the moving car, the fan seems to spin only, even though we know the car is moving too. 
• This idea of breaking down the motion of things into spinning and straight-line movements is really useful when studying how things move.
• Studying the motion of objects as a combination of rotational and translational motion is nothing but studying rolling motion.
• While studying rolling motion, we assume the translation of the centre of mass and rotation of the rest of the object around the centre of mass. 

Rolling Motion

1. In the image (a), pure rotation is shown. Every point on the wheel rotates about the centre of mass with angular speed ω. And every point on the outside edge of the edge of the wheel has linear speed V_COM = ωR, where R is the radius of the wheel. 

2. In the image (b), pure translation is shown. Every point on the outside edge of the wheel moves to the right with linear speed V_COM.

3. In the image (c), actual rolling motion is depicted.

Kinetic Energy of Rolling Motion

Consider a disc of radius R rolling over a horizontal smooth surface without slipping. The wheel has two types of motion: (i) linear motion of COM with velocity VCM along the horizontal and (ii) rotational motion of the wheel about the horizontal axis passing through its centre, with an angular speed ω. 

M → Mass of  wheel
V_CM → Velocity of the centre of mass
Linear kinetic energy of the centre of mass, K_T= ½ MV_CM²
"I" → Moment of inertia
"ω" Angular velocity
Rotational kinetic energy, K_R= ½ Iω² 

When a system of particles exhibits simultaneous translational and rotational motion the kinetic energy of a system of particles, KE, can be written as the sum of the kinetic energy due to translation, and kinetic energy due to rotation: 

KE = K_T + K_R   =  ½ MV_CM² + ½ Iω²

V_CM=Rω so ω=V_CM/R and I= Mk², where k is the radius of gyration of the body

Putting the value of ω and I in the formula of kinetic energy we get, 

KE=  ½ MV_CM² + ½ Mk²(V_CM/R)² 

KE=  ½ MV_CM² [1+(k²/R²)]

So, if we know the moment of inertia or radius of gyration of a body, then we can calculate the kinetic energy of that object.

Special Cases:

Solved Examples

Q1: A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations
respectively sph cyl (Ksph :Kcyl ) is 2√x. The value of X is.   [ JEE Main 2023 (Online)]
Ans: 5
Let the mass of both the solid sphere and the solid cylinder be M, and let their common radius be R. The moment of inertia I of a solid sphere and a solid cylinder are given by:

The radius of gyration k is related to the moment of inertia I by the formula I = Mk². Therefore, we can find the radius of gyration for both the solid sphere and the solid cylinder using their respective moments of inertia:

Now, let's find the ratio of their radius of gyrations:

Squaring both sides:

Substituting the expressions for

Simplifying and solving for x:

Q2: For a rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is x/5. The value of X is ___________.      [JEE Main 2023 (Online)]

Ans: 2
Sol: For a rolling spherical shell, we must consider the fact that it has both translational and rotational kinetic energy. The total kinetic energy (K_total) can be expressed as the sum of the translational kinetic energy ( K_trans) and the rotational kinetic energy (K_rot).

The translational kinetic energy of an object with mass (m) and linear velocity (v) is given by:

The rotational kinetic energy of a rolling spherical shell with moment of inertia (I) and angular velocity (ω) is given by:

For a rolling object without slipping, the relationship between linear velocity (v) and angular velocity (ω) is:
v = Rω

Where R is the radius of the spherical shell.
The moment of inertia for a spherical shell is given by:

Now, we can substitute the moment of inertia and the relationship between linear and angular velocity into the equation for rotational kinetic energy:

Now, we can find the ratio of rotational kinetic energy to total kinetic energy:

Multiplying both the numerator and the denominator by 6:

Comparing this to the given ratio of x/5, we can determine that the value of x is 2.