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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.

 

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 {{\rm{\vec r}}_{\rm{1}}},\,{{\rm{\vec r}}_{\rm{2}}},..........,{{\rm{\vec r}}_{\rm{n}}} as shown in the figure, the position vector for the center of mass {{\rm{\vec r}}_{{\rm{cm}}}} is  defined as

{{\rm{\vec r}}_{{\rm{cm}}}} = {{{m_1}{{{\rm{\vec r}}}_{\rm{1}}} + {m_2}{{{\rm{\vec r}}}_{\rm{2}}} + ........ + {m_n}{{{\rm{\vec r}}}_{\rm{n}}}} \over {{m_1} + {m_2} + ........... + {m_n}}}

or {{\rm{\vec r}}_{{\rm{cm}}}} = {{\sum {{m_i}{{{\rm{\vec r}}}_{\rm{i}}}} } \over M}

where M = \sum {{m_i}} is the total mass of the system.
The components of the above equation may be written as

{x_{cm}} = {{\sum {{m_i}{x_i}} } \over M}, {y_{cm}} = {{\sum {{m_i}{y_i}} } \over M}, {z_{cm}} = {{\sum {{m_i}{z_i}} } \over M}

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

{{\rm{\vec r}}_{{\rm{cm}}}} = {1 \over M}\int {{\rm{\vec r}}dm}

where {\rm{\vec r}} is the position vector of the centre of mass of a small mass element dm.
The components of this equation are

{x_{cm}} = {1 \over M}\int {xdm} , {y_{cm}} = {1 \over M}\int {ydm} and {z_{cm}} = {1 \over M}\int {zdm}

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 xcm = 0.

In this case it is convenient to express the mass element in terms of the angle \theta , measured in radians. The element, which subtends an angle {d_\theta } at the origin, has a length R{d_\theta } and a mass dm = \lambda R{d_\theta }. (l = mass per unit length)
Its y coordinate is y = R\sin \theta .

Therefore, {y_{cm}} = {1 \over M}\int {ydm}  takes the form
{y_{cm}} = {1 \over M}\int\limits_0^\pi {\lambda {R^2}\sin \theta d\theta } = {{\lambda {R^2}} \over M}\left[ { - \cos \theta } \right]_0^\pi = {{2\lambda {R^2}} \over M}
The total mass of the ring is M = \pi R\lambda ; therefore, {y_{cm}} = {{2R} \over \pi }

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 m1, m2, ……., mn and whose total mass is M which is a constant. From the definition of the centre of mass, we have

M{{\rm{\vec r}}_{{\rm{cm}}}}  = m1{{\rm{\vec r}}_{\rm{1}}} + m2{{\rm{\vec r}}_{\rm{2}}} +………………+ mn{{\rm{\vec r}}_{\rm{n}}}

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

M{{\rm{\vec v}}_{{\rm{cm}}}} = {m_1}{{\rm{\vec v}}_{\rm{1}}} + {m_2}{{\rm{\vec v}}_{\rm{2}}} + ......... + {m_n}{{\rm{\vec v}}_{\rm{n}}}

where {{\rm{\vec v}}_{\rm{1}}}\left( { = {{d{{{\rm{\vec r}}}_{\rm{1}}}} \over {dt}}} \right) is the velocity of the first particle, etc., and {{\rm{\vec v}}_{{\rm{cm}}}}\left( { = {{d{{{\rm{\vec r}}}_{{\rm{cm}}}}} \over {dt}}} \right) is the velocity of the centre of mass.

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

M{{\rm{\vec a}}_{{\rm{cm}}}} = {m_1}{{\rm{\vec a}}_{\rm{1}}} + {m_2}{{\rm{\vec a}}_{\rm{2}}} + ......... + {m_n}{{\rm{\vec a}}_{\rm{n}}}

where {{\rm{\vec a}}_{\rm{1}}} is the acceleration of the first particle, etc., and {{\rm{\vec a}}_{{\rm{cm}}}} is the acceleration of the centre of mass.  Now, from Newton’s second law the force {{\rm{\vec F}}_{\rm{1}}} acting on the first particle is given by {{\rm{\vec F}}_{\rm{1}}} = {m_1}{{\rm{\vec a}}_{\rm{1}}}. Likewise, {{\rm{\vec F}}_2} = {m_2}{{\rm{\vec a}}_2}, etc. We can then write the above equation as

M{{\rm{\vec a}}_{{\rm{cm}}}} = {{\rm{\vec F}}_{\rm{1}}} + {{\rm{\vec F}}_{\rm{2}}} + ......... + {{\rm{\vec F}}_{\rm{n}}}

Among all these forces the internal forces are exerted by the particles on each other. However, from Newton’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

M{{\rm{\vec a}}_{{\rm{cm}}}} = \sum {{{\overrightarrow F }_{external}}}

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 \sum {{{{\rm{\vec F}}}_{{\rm{external}}}}} = \overrightarrow 0 then {{\rm{\vec v}}_{{\rm{cm}}}} = 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).  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 keep  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 = {{10 \times 20 + 40.x} \over {10 + 40}}

The final  x-coordinate of the centre of mass
= {{10\left( {20 - 8 + d} \right) + 40\left( {x + d} \right)} \over {10 + 40}}

Equating the two, we get
d={8 \over 5}m
The  dog is 20 – 8 + {8 \over 5} = 13.6 m from the shore.

Application 3

Two balls with masses m1 = 3 kg and m2 = 5 kg have initial velocities v1 = v2 = 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 vcm is constant for all times.
From equation,  in component form

{v_{cmx}} = {{{m_1}{v_{1x}} + {m_2}{v_{2x}}} \over M} = {{\left( {3kg} \right)\left( { - 5\cos {{37}^o}m/s} \right) + \left( {5kg} \right)\left( {0m/s} \right)} \over {8kg}} = - 1.5m/s

{v_{cmy}} = {{{m_1}{v_{1y}} + {m_2}{v_{2y}}} \over M} = {{\left( {3kg} \right)\left( { - 5\sin {{37}^o}m/s} \right) + \left( {5kg} \right)\left( {5m/s} \right)} \over {8kg}} = + 2m/s

Thus, {{\rm{\vec v}}_{{\rm{cm}}}}= -1.5\hat i+ 2\hat jm/s

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

{{\rm{\vec r}}_{{\rm{cm}}}}={{\rm{\vec v}}_{{\rm{cm}}}}t = -3\hat i+ 4\hat jm

Application 4

 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.

{{\rm{\vec v}}_{{\rm{PG}}}} = {v_p}\hat i

The velocity of man relative to the ground is then

{{\rm{\vec v}}_{{\rm{MG}}}} = {{\rm{\vec v}}_{{\rm{MP}}}} + {{\rm{\vec v}}_{{\rm{PG}}}} = (2 + {v_p})\hat i

From the conservation of momentum, we have
(75 + 25)  \times 4 = 75( 2 + vp) + 25 vP
Thus, the velocity of the platform is vP = 2.5 m/s, and the velocity of the man

is vm = 4.5 m/s.

Since the velocity of man relative to platform is 2m/s, therefore, it takes  t = {{4m} \over {2m/s}} = 2s for him to walk from the rear to the front.

Displacement of platform is               xp = vpt = (2.5)(2) = 5m
Displacement of the man is               xm = vmt = (4.5)(2) = 9m
Displacement of the center of mass   xcm = vcmt = (4)(2) = 8m



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