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Rolling motion is formed due to two types of motion, or it is a combination of two types of motion, which are translational motion and rotational motion. The translational motion of a body or object is the movement of the centre of mass. The surfaces in contact are deformed temporarily during the rolling motion of a body. Some examples of rolling motion include the movement of a car wheel or a ball’s motion along the ground. The two types of motions responsible to make Rolling motion are:
- Translatory motion: Translatory motion is a motion where all portions of the body are seen to move the same distance in the duration. Translational motion is of two types: Rectilinear motion and Curvilinear motion.
- Rotational motion: Rotational motion can be expressed as a type of motion found in an object that moves around in a circular path, for a fixed orbit. Some instances of the same include the motion of the earth about its own axis, the motion of motors or gears, etc.
Table of Content |
Key Terms: Rolling Motion, Center of Mass, Rotational Motion, Translatory Motion, Velocity, Rectilinear Motion, Moment of Inertia, Rigid Body, Kinetic Energy
What is Rolling Motion?
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Any circular body, like a sphere or wheel, undergoes Rolling motion as it passes over a horizontal surface. Rolling motion examples are a ball’s motion along the ground. At every instant of rolling, there occurs a single point of contact. Rolling motion can be defined as:
“Rolling Motion combines the rotational and translational motion of an object with respect to a surface, such that, in case ideal conditions exist, the two bodies keep being in contact with one another without sliding.” |
Rolling Motion
Imagine a circular body rolling over a surface continuously, without any obstructions. That body, at all times, is in contact with the surface. Simply put, at any particular instant of time, the area of the circular body in contact with the surface is in a state of rest with respect to the surface.
Translational and Rotational Motion
Rolling Motion is a combination of both Translational Motion and Rotational Motion.
- The motion of the centre of mass is the translational motion of the body.
- When an object is in Rolling Motion, the surfaces in contact face temporary deformation.
- This causes a finite area of both bodies to come in contact with each other.
- The component of contact force parallel to the surface opposing motion leads to the creation of friction.
Thus, it can be shown using the following example:
Rolling Motion of Disc:
Rolling Motion of Disc
- Vcm is the velocity of the centre of mass of a circular body.
- The centre of mass lies at the geometric centre C, the velocity of C is thus Vcm (parallel to the rolling surface).
- Rotational motion occurs at the axis of symmetry.
- Thus, the velocity at any point P0, P1, or P2 consists of two parts: Translational Velocity Vcm and Linear Velocity Vr caused due to rotational motion.
Thus, the rolling motion formula is,
vr = rω |
Here,
- ω = angular velocity of circular body.
- vr = perpendicular to radius vector at any point lying on the disc with respect to geometric centre C.
The kinetic energy of such a rolling body is given by the sum of the kinetic energies of translational motion and rotation.
Hence,
Where,
- m is the mass of the body
- vcm is the rotational motion
- I is the Moment of Inertia
- ω is the angular velocity of the rolling body
Rolling Motion
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Point of Contact
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The area of the rolling object that is in contact with the surface is in a position of momentary rest. At any given moment, any point that is in contact with the surface is at rest as the body continues rolling.
Pure Rolling Motion can be observed as either the combination of translational motion and rotational motion about the center of mass or the momentary rotational motion about the point of contact.
Point of Contact
As the point of contact is at momentary rest in pure rolling, its resultant velocity v is zero (v = 0). At the point of contact, vTRANS is forward (to right) and vROT is backwards (to the left).
Kinetic Energy for Rolling Motion
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Since rolling is known to be a combination of rotational motion and translational motion. Thus, it can be said that:
⇒ K.E = KT + KR
= \(\frac{1}{2}\)mv² + \(\frac{1}{2} \) Iω²
= \(\frac{1}{2} \)m vcm² + \(\frac{1}{2} \)Iω²
As we know, ‘I’ is the moment of inertia. Thus, we can also write the moment of inertia in the form of I = mk², k = radius of gyration.
= \(\frac{1}{2} \)m vcm² + \(\frac{1}{2} \)mk² ω²
In the case of rolling without slipping, the mathematical condition that it should meet is rω = vcm
K = \(\frac{1}{2} \)m² ω² + \(\frac{1}{2} \)mk² ω²
= \(\frac{1}{2} \) m vcm² + \(\frac{1}{2} \)mk² \(\frac{v_{cm^2}}{r^2}\)
= K = \(\frac{1}{2} \)m vcm² [ 1 + \(\frac{k²}{r²} \) ]
Types of Motion of Rigid Body
Some of the types of motion of a rigid body include:
- Assume a solid metallic cylinder rolling down the same slanted plane. The cylinder is known to shift from the top to the bottom of the inclined plane, indicating translational motion. Yet, all the particles are not travelling at the same speed at a given period of time. Consequently, the body isn’t moving in a true translational motion. Thus, it is a combination of translational and rotational motion.
- In case a rigid body is not pivoted in any way, its motion is either considered a pure translation or a combination of translation and rotation. The motion of a rigid body pivoted in some way is called rotation. The rotation can be fixed, such as with a ceiling fan.
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Things to Remember
- Rolling motion is a type of motion formed due to two types of motion, which are translational motion and rotational motion.
- The kinetic energy for rolling motion is K = \(\frac{1}{2} \)m vcm² [ 1 + \(\frac{k²}{r²} \) ].
- The formula of rolling motion is vr = rω.
- Translatory motion can be defined as a type of motion where all portions of the body move the same distance for a duration of time.
- Rotational motion is a type of motion where an object is moving around a circular path, in a fixed orbit.
Previous Year Questions
- A thin circular plate of mass M and radius R … [JEE Mains 2019]
- A rectangular solid box of length … [JEE Mains 2019]
- A string is wound around a hollow cylinder of mass … [JEE Mains 2019]
- A tennis ball (treated as hollow spherical shell) … [JEE Mains 2013]
- A rigid massless rod of length … [JEE Mains 2019]
- A roller is made by joining together … [JEE Mains 2016]
- A slender uniform rod of mass … [JEE Mains 2017]
- A solid sphere of mass M and radius R … [JEE Mains 2019]
- A thin bar of length L has a mass per unit length … [JEE Mains 2014]
- A thin disc of mass MM and radius RR has mass per … [JEE Mains 2019]
- A uniform rectangular thin sheet … [JEE Mains 2019]
Sample Questions
Ques. What is Rolling Motion definition? (1 mark)
Ans. Rolling motion is composed of two motions, Translational motion and Rotational motion. While the translational motion of an object is the movement of the centre of mass, Rotational motion can be found when surfaces in contact are deformed temporarily during the rolling motion of a body.
Ques. What is Translatory Motion? Explain with example. (1 mark)
Ans. When an object or a body is seen to move from one place to another with respect to the frame to reference, it is then known as translatory motion. Example of Translatory Motion: Motion of a car or bus in a straight highway or road.
Ques. Is it possible for a sphere to roll purely on a smooth inclined surface? (2 marks)
Which is to say, ‘ω’ should simultaneously also increase. It needs torque, which in this case, can be offered by friction. Hence, pure rolling on an incline needs the surface to be rough.
Ques. S1 and S2 are two spheres which have equivalent masses. Here, S1 has been found to roll down a smooth inclined plane of length 5 cm, and height 4 cm. Then, S2 is seen to fall vertically down by 4 cm. Determine whether the work was done by all forces which acted on S1 and on S2 are the same and non-zero. (1 mark)
Ans. By means of energy balance, it can be found that potential energy that depends on height changes to kinetic energy at the bottom. This means the work done in either case is the same. The work done would differ in case friction was present.
Ques. What is Rotational Motion? (1 mark)
Ans. In a fixed orbit, Rotational motion is expressed as the motion of an object around a circular path.
Ques. Determine the displacement of the centre of the wheel in one rotation. Here, assume the radius of the wheel to be R. (2 marks)
Ans. Consider the time taken for one rotation = ‘T’.
Now let angular speed of the wheel be, w = 2π/T
The velocity of the centre is Rw = R * 2π/T.
∴ The Distance moved = R * 2π/T * T = 2πR.
Ques. Examine the moment of inertia of a rod whose mass is 30 kg and altitude is 30 cm? (2 marks)
Ans. The parallel axis theorem formula for a rod is given as,
I = (1/12) mr2
putting in the values we get
I = 0.225 Kg m2.
Ques. Evaluate the moment of inertia of a stick whose mass is 100 gm and size is 10 cm? (2 marks)
Ans. According to the parallel axis theorem procedure for a rod is given as-
I = (1/12) mr2
putting in the values we get-
I = 0.0000833 Kg m2.
Ques. (a) Find the moment of inertia of a sphere about a tangent to the sphere. It is given that the moment of inertia of the sphere about any of its diameters to be 2 MR2/5, where M is considered as the mass of the sphere and while R is considered to be the radius of the sphere.
(b) The moment of inertia given a disc of mass M and radius R about any of its diameters to 1 is 1/4 MR2. Now Find the moment of inertia of that disc about an axis normal to the disc moving and going through a point on its edge. (3 marks)
Ans. (a) It is given that the Moment of inertia of the given sphere about any diameter is 2/5 MR2.
Now by applying the parallel axis theorem, the Moment of inertia of a sphere about a tangent to the sphere is
2/5 MR2 +M(R)2 =7/5 MR2
(b) It is given that the moment of inertia of the given disc about any of its diameters is 1/4 MR2
(i) To find the solution we are using the Perpendicular axis theorem.
So by applying this theorem the moment of inertia of the disc about an axis moving through its centre and normal to the disc is
2 x 1/4 MR2 = 1/2 MR2.
(ii) Now by applying the parallel axis theorem we can find the moment of inertia.
Thus, the moment of inertia of the disc moving through a point on its edge and normal to the disc is
= 1/2 MR2+ MR2
= 3/2 MR2.
Ques. Consider a rope of bare mass applied around a hollow cylinder of mass given 3 kg and radius considered to be 40 cm. Find out the angular acceleration of the cylinder if given that the rope is ripped with a force of 30 N? Also find the linear acceleration of the rope? Note- Assume that there is no slipping. (5 marks)
Ans. Here it is given in the question that the M = 3 kg, R = 40 cm = 0.4 m
By applying the parallel axis theorem we can find the moment of inertia.
The Moment of inertia of the hollow cylinder about its axis is
I = MR2 = 3(0.4)2 = 0.48 kg m2
Here it is given that the force applied is
F = 30N
Torque = F×R.
= 30×0.4
= 12N-m
Now if the “a” angular motion then
a= torque÷moment of inertia
a = 12÷0.48
a= 25 rad s-2
Now the linear acceleration is
a = Ra = 0.4×25 = 10ms-2
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