# Chapter Notes: Rotational Motion and Rolling Motion Physics Class 11

Notes for Rotational Motion chapter of class 11 physics. Dronstudy provides free comprehensive chapterwise class 11 physics notes with proper images & diagram.

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**CENTRE OF MASS**

In a system of extended bodies there is one special point that has some interesting and simple properties no matter how complicated the system is. This point is called the ** center of mass**.

For a system of n particles whose position vectors are as shown in the figure, the position vector for the center of mass isÂ defined as

or

where is the total mass of the system.

The components of the above equation may be written as

, ,

The location of the centre of mass is independent of the reference frame used to locate it. The centre of mass of the system of particles depends only on the masses of the particles and the positions of the particles relative to one another.

A rigid body, such as a meter stick, can be thought of as a system of closely packed particles. Hence it also has the centre of mass. The number of particles in the body is so large and their spacing so small, however, that we can treat the body as though it has a continuous distribution of mass. For continuous distribution of mass, the centre of mass is defined as

where is the position vector of the centre of mass of a small mass element *dm.
*The components of this equation are

, and

Often we deal with homogeneous objects having a point, a line, or a plane of symmetry. Then the centre of mass will lie at the point, on the line, or in the plane of symmetry. For example, the centre of mass of a homogeneous rod will lie at the centre of the rod, the centre of mass of a homogenous sphere will be at the centre of the sphere, the centre of mass of a cone will be at the axis of the cone, etc.

**Application 1**

**Find the center of mass of a uniform semicircular rod of radius R.**

**Solution**

From the symmetry of the body we see at once that the centre of mass must lie along the *y *axis, so *x _{cm} *= 0.

In this case it is convenient to express the mass element in terms of the angle , measured in radians. The element, which subtends an angle Â at the origin, has a length Â and a mass . (l = mass per unit length)

Its y coordinate isÂ .

Therefore, Â takes the form

The total mass of the ring is ; therefore,

**Motion of the Centre Of Mass**

Now we can discuss the physical importance of the centre of mass concept. Consider the motion of a group of particles *m*_{1}, *m*_{2}, â€¦â€¦., *m*_{n} and whose total mass is *M *which is a constant. From the definition of the centre of mass, we have

*M* Â = *m*_{1{{\rm{\vec r}}_{\rm{1}}} + }*m*_{2{{\rm{\vec r}}_{\rm{2}}} +â€¦â€¦â€¦â€¦â€¦â€¦+ }*m*_{n{{\rm{\vec r}}_{\rm{n}}}}

Differentiating this equation w.r.t. time, we obtain

where Â is the velocity of the first particle, etc., and Â is the velocity of the centre of mass.

Differentiating the above equation w.r.t. time, we obtain

where is the acceleration of the first particle, etc., and is the acceleration of the centre of mass.Â Now, from Newtonâ€™s second law the force acting on the first particle is given by . Likewise, , etc. We can then write the above equation as

Among all these forces the internal forces are exerted by the particles on each other. However, from Newtonâ€™s third law, these internal forces will occur in equal and opposite pairs, so that they contribute nothing to the sum. The right hand sum in the above equation then represents the sum of only the *external *forces acting on all the particles (system). We can then rewrite the above equation as

This states that the centre of mass of a system of particles moves as though all the mass of the system is concentrated at the centre of mass and all the external forces were applied at that point.

One important situation is that in which then = constant.

**Application ****2**

**A dog of mass 10 kg stands on a stationary boat of mass 40 kg so that he is 20 m from the shore. He then walks 8 m on the boat towards the shore and halts. How far is he from the shore now? Assume that there is no friction between the boat and water. **

**Solution**

In the horizontal direction, there is no force on the system (dog + boat).Â Therefore, the centre of mass of the system does not move in the horizontal direction. Since the dog moves towards the shore, the boat moves away from the shore to keepÂ centre of mass stationary. Let *d *be the distance by which the boat moves backwards and let *x *be the initial distance of the boat from the shore.

The initial *x*-coordinate of the centre of mass =

The finalÂ x-coordinate of the centre of mass

=

Equating the two, we get

d=m

TheÂ dog is 20 â€“ 8 + = 13.6 m from the shore.

**Application 3**

**Two balls with masses m _{1} = 3 kg and m_{2} = 5 kg have initial velocities v_{1} = v_{2} = 5 m/s in the directions as shown in the figure. They collide at the origin.**

**(a) Find the velocity of the center of mass 3 s before the collision**

**(b) Find the position of the center of mass 2 s after the collision**

**Solution**

(a) The given time is of no consequence since *v _{cm }*is constant for all times.

From equation,Â in component form

m/s

m/s

Thus, = -1.5+ 2m/s

(b) Since the collision occurs at the origin, the position of the center of massÂ 2s later is

=*t* = -3+ 4m

**Application 4**

Â **A 75 kg man stands at the rear end of a platform of mass 25 kg and length 4m, which moves initially at 4 m/s over a frictionless surface. At t = 0, he walks at 2 m/s relative to the platform and then stops at the front end. During the period of walking, find the displacement of
**

**(a) the platform,**

**(b) the man,**

**(c) the center of mass,**

**Solution**

Initially the man, the platform, and the *cm *have the same velocity, 4m/s. When he begins to walk forward, his increase in momentum must be compensated by a decrease in the platform's momentum. Let us say that the velocity of the platform relative to the ground while he is walking i.e.

The velocity of man relative to the ground is then

From the conservation of momentum, we have

(75 + 25) 4 = 75( 2 + *v _{p}*) + 25

*v*Thus, the velocity of the platform is

_{P }*v*= 2.5 m/s, and the velocity of the man

_{P}is

*v*= 4.5 m/s.

_{m}Since the velocity of man relative to platform is 2m/s, therefore, it takes Â Â for him to walk from the rear to the front.

Displacement of platform is Â Â Â Â Â Â Â Â Â Â Â Â Â Â *x _{p} *=

*v*= (2.5)(2) = 5m

_{p}tDisplacement of the man is Â Â Â Â Â Â Â Â Â Â Â Â Â

*x*=

_{m}*v*= (4.5)(2) = 9m

_{m}tDisplacement of the center of mass Â Â

*x*=

_{cm}*v*= (4)(2) = 8m

_{cm}t