Puzzle of Centre of Mass Solved

Centre of Mass: Centre of mass is a physical concept which is used to simplify the analysis of motion of a system (continuous or discrete). It is in general defined as a point where the whole mass of the system can be assumed to be concentrated. Generally the concept of centre of mass is used wherever the distribution of mass of the system is uniform and some kind of symmetry exists in the system as will be discussed later.

Consider a uniform spherical ball. Let it be thrown upwards by a spinner who also spins the ball. Thus, the ball goes spinning upwards. Now consider the motion of a point on the surface of the ball. This point rotates in a circular motion as well as moves upwards. Thus, its motion is complicated to analyze. But consider the motion of centre of the ball. The centre of the ball executes a straight line trajectory and hence its motion is much simpler to analyze.

It will be discussed later that centre of mass of such a symmetrical ball will lie on its geometrical centre and thus motion of centre of mass is much simpler to analyze.

Position or Coordinates of Centre of Mass: In order to find the position of centre of mass of a system, it is necessary to have the knowledge of the position of various masses in case of a discrete system or the distribution of mass in case of a continuous system.

C.M. of a Discrete System : Consider the collection of N particles. Let mass of i^th particle be m_i and its x_i. Then we write the product m_ix_i for each particle and add then to obtain
and similarly and

Then lets define

Where, is the total mass of the system. Locate the point, this point is called the centre of mass of the system.

Thus, if position vector of i^th particle is then centre of mass is defined to have position

NOTE : Centre of mass is also defined in this way.

Centre of Mass of Continuous System :

Consider a body having a continuous distribution of mass. Consider a small element of the body having a mass dm. If (x, y, z) are the coordinates of this small mass dm, then coordinates of c.m. of the system may be written as

Thus, for a continuous system of masses.

The limits of integration are chosen in a way that as the integration variable goes through the limits, the elements cover the entire body.

Examples of Discrete System :

Q. 1 There particles of masses m₁ = 0.50 kg, m₂ = 1.0 kg and m₃ = 1.5 kg are placed at three corners of a right angled triangle of sides 3.0 cm, 4.0 cm and 5.0 cm as shown in figure locate the c.m. of the system.

Solution : In order to locate the position of the centre of mass it is important to choose a particular origin w.r.t. which the position of the c.m. is to be specified.

Let the mass of 0.5 kg be the origin and the 3.0 cm line and 4.0 cm line act as y and X-axes respectively. Then the position coordinates of various masses are as below.

mass    X-word    Y-word

0.50 kg    0    0

1.0kg    4.0 cm    0

1.5 kg    0    3.0 cm

Thus, x-coordinate of center of mass :

y-coordinate of center of mass :

Thus, c.m. of the system is located at (1.3, 1.5) cm w.r.t. the chosen origin.

Q. 2. A circular plate of diameter d is kept in contact with a square plate of edge d as shown below. The density of the material and thickness are same everywhere. Locate the position of the centre of mass.

Solution : In order to locate the centre of mass of the above discrete system we neeed to first know the position of the c.m. of the two bodies individually.

By symmetry c.m. of the circular plate and the square plate both will coincide with their geometrical centres i.e., O and P respectively.

Let the origin of the system be located at the ground on the line containing the c.m. of the circular plate. Then the position of the c.m. of two bodies can be given as.

Mass    x-word    y-word

m₁ (mass of circular plate)    0    d/2

m₂ (mass of square plate)    d    d/2

Let, the density of material of both plates be and thickness a << d (negligible thickness), then

Mass = density × area

Thus, mass of circular plate    

mass of square plate        

Thus, x-coord. of c.m.,    

The above result is also based on a very important result that if we know the centres of masses of parts of the system and their masses, we can get the combined c.m. by treating the parts as point particles placed at their respective centres of mass.

Before taking examples of a continuous distribution of masses it is important to discuss the role of symmetry of the body and uniform distribution of mass in determination of the centre of mass (especially of continuous bodes).

Symmetry and Centre of Mass : In general a continuous three dimensional body can possess the following three types of symmetry :

(i) Centre of Symmetry (Inversion Symmetry): A body is said to possess a centre of symmetry when for every point having a position vector w.r.t. O there exists an identical point having a position vector w.r.t. O.

eq. sphere (centre of sphere), cube (center of cube) etc.

Whenever a body possesses a centre of symmetry and has a uniform distribution of mass (, density is constant) then the centre of mass of the body coincides with the centre of symmetry.

(ii) Line of Symmetry (Rotational Symmetry) : A body is said to possess a line of symmetry if on rotating the body about this line doesn’t produce any change in the configuration of the body.

Whenever a body possesses a line of symmetry and has a uniform distribution of mass then the centre of mass of the body lies on the line of symmetry e.g. uniform ring (line through centre), uniform sphere (Line through centre) etc.

Note that a body possessing a centre of symmetry necessarily possesses a line of symmetry but the inverse is not always true e.g. considers a uniform semicircular ring. It possesses a line of symmetry xx¹ but not a centre of symmetry. Thus in order to obtain its centre of mass if we consider O as the origin we need not bother about the x-component of the c.m. as it will lie on the line xx¹ and its x-component is necessarily O.

(iii) Plane of Symmetry (Reflection Symmetry) : A body is said to possess a plane of symmetry if on reflection of the body through this plane the configuration of the body is exactly identical and the two images are mirror images of each other. If a body possesses a plane of symmetry and has a uniform distribution of mass then the c.m. Lies on the plane of symmetry.

Note that the condition of uniform distribution of mass i.e., a constant density is sufficient e.g. if the density of the sphere isn’t constant then the centre of mass doesn’t necessarily coincides with the geometrical centre of the body (sphere).

Examples of Continuous System :

(i) Centre of mass of a uniform Straight Rod :

Consider a rod of length L and mass M possessing a continuous distribution of mass and hence a constant density . Let the rod be placed such that its one end coincides with the origin and other and other end at a distance L from origin along x-axis. Since rod is uniform

Now, consider a small element of thickness dx on the rod at a distance x from the origin.

Mass of the small element    

Thus, the x-coordinate of c.m.    

Y coord. of c.m.        as y coord. of the small element is O. Similarly

Thus, the coordinate of centre of mass are

(ii) Centre of Mass of a uniform semicircular wire :

Consider a uniform semicircular wire of radius R and Mass m with a constant density of mass ().

Consider a small element of the is wire with thickness Rd as shown above.

Hence mass of the small element

Hence from the definition of the centre of mass, coordinates of the c.m. are given by

Now, for the small element the x and y coord. given by (R cos , R sin )

= 0

As already discussed the above result can also be directly obtained using arguments of symmetry (line of symmetry)

Since the wire is planer with no distribution of mass along z-axis, therefore z coord. of c.m. is O.

Thus, coordinates of the centre of mass are.

(iii) Centre of mass of a uniform semicircular plate.

Consider a semicircular disc of mass M and radius R. (Let thickness of the disc be negligible). Then the semicircular disc can be visualized as a number of semicircular wires of constant density  piled over each other to form the disc.

The problem can be solved using the result that any part of the system may be replaced by a point particle of the same mass placed at the centre of mass of that part.

Consider are such ring of radius r and thickness dr.

Since the disc is uniform, mass per unit area of the disc

Area of the small ring so considered

Thus, the mass of the ring    

Now, the coordinates of the c.m. of this ring as obtained were

Thus, the coordinates of the c.m. of the disc are

Thus, coordinates of c.m. of disc are

Motion of the Centre of Mass :

Consider a system of n particles, the i^th particle having mass m_i and position vector with respect to an inertial frame. Each particle is acted upon by forces due to all other (N–1) particles and external forces due to sources outside the system.

Acceleration of the i^th particle is the thus,

{From Newton’s second law}

…(i)

Writing similar equations for all the N particles of the system and adding them up.

…(2)

Now, the term represents the force on the i^th particle due to j^th particle and by Newton’s third law of motion the i^th particle applies an equal and opposite force on the j^th particle, therefore the sum over all the particles for such a force will be O i.e.,

is the vector sum of all the external forces acting on the system of N particles.

…(3)

Also, from the definition of the centre of mass

Differentiating the above expression twice w.r.t. time.

…(4)

where = acceleration of the i^th particle and = acceleration of the centre of mass of the system.

from (3) and (4) :

…(*)

Above equations gives a very important result that whenever the vector sum of all forces acting on a system is zero, the acceleration of the centre of mass of the system is O i.e.,

Thus, if is velocity of the centre of mass, then

If initially the centre of mass of the system was at rest w.r.t. some inertial frame of reference, it will continue to be at rest w.r.t. that frame. The individual particles may go on complicated paths changing their positions but the centre of mass will be obtained at the same position.

When external force do not add up to zero the centre of mass is accelerated and acceleration is given by the equation.

Thus, if instead of the system we have an individual particle of mass M placed at the position of the centre of mass then the acceleration produced in it would be the same as that of the whole system.

Problem 1 : Consider two uniform spheres of masses m and 2M initially placed at a distance ‘d‘ apart. They exert gravitational attraction on each other. Calculate the distance moved by the two with respect to their initial positions when they actually collide with each other.

Solution : First of all by arguments of symmetry the centre of masses of both the spheres coincide with the centre of the spheres. Thus, we can replace both the spheres by point masses at their centres.

Now, if we consider the system to consist of both the masses, then the gravitational attraction exerted by them on each other will constitute internal forces. Since no external forces act on this system the acceleration of the c.m. of this system will have no acceleration, and thus its velocity will remains constant.

Since initially the two masses were at rest and this centre of mass was also at rest, it will remain at rest thought the two masses will move towards each other.

At the time of collision both the particles will be the same place and the c.m. will also be there. As the c.m doesn’t move collision takes place at the centre of mass.

Taking the origin of the system at the centre of sphere of mass M, the initial position of the c.m. is.

Thus, collision will take place at a distance from the sphere of mass M.

Distance moved by mass    

Distance moved by mass    

Problem 2 : A cubical block of ice of mass m and edge L is placed in a large tray of mass M. If ice melts, how far does the centre of mass of the system (ice plus tray) comes down?

Consider the origin of the system at the point O. Let the c.m. of the tray be at height x₁ from origin and that of cube distance x₂.

Initial coordinates of the c.m. are

…(1)

Now as the cube of ice is acted upon by external agencies (heat) it melts and its centre of mass comes down. Since the tray is large the height of water is negligible in the tray and thus the new position of the centre of mass is

Thus, the new position of c.m. of the system is

Thus, the shift in the c.m.    

Note that in the above problem we assumed that as heat is observed by the ice cube it melts uniformly on the tray so that the horizontal position of its center of mass doesn’t change.

Also in the above problem, an external force of gravity acts on the cube and the tray throughout but it is balanced by the force of normal reaction when the cube is in solid form and when the molecular forces of attraction are not able to hold it together gravity causes the c.m. to fall down.

Problem : 3 A block of mass M is placed on the top of bigger block of mass 1OM. All the surfaces are friction less. The system is released from rest. Find the distance moved by the bigger block at the instant the smaller block reaches the ground.

Solution : When the system is released from rest, a vertical force of gravity acts on the system. But no horizontal force is there. Thus, the x-coordinate of the centre of mass of the system should not change as initially it was at rest.

If the bigger block moves towards right with a distance x, the smaller block will move towards left by a distance (2.2 m – x). Taking the initial position of the c.m. to be at the origin.

Problem 4 : The balloon, the light rope and the monkey shown in figure are at rest in the air. If the monkey reaches the top of the rope, by what distance does the balloon descend? Mass of balloon = M, mass of monkey cm and length of rope ascended by monkey = L

Since, initially the balloon, rope and monkey were at rest in air, the resultant of forces acting on them as a system was equal to 0 and the centre of mass of the system was at rest with position coordinates given by

where the origin is taken at the position of the monkey.

Now as the monkey starts ascending up the rope it will exert a tension T in the rope due to its weight mg and acceleration. But all these forces are internal to the system and thus should not affect the position of the c.m.

Since, the monkey reaches top of rope Let it ascends a distance x, Let the balloon descend by distance then, coord. of c.m. will remain intact, thus monkey and balloon will meet at the position of c.m. thus, distance ascended by balloon

Problem 5 : man of mass M having a bag of mass m slips from the roof of a tall building of height H and starts falling down vertically. When at height h from the ground he notices that ground below him is pretty hard but there is a pond at a horizontal distance x from the line of fall. In order to save himself he throws the bag horizontally (with respect to himself) in the direction opposite to the pond. Calculate the minimum horizontal velocity imparted to the bag so that the man lards in water. If the man just succeeds to avoid hard ground, where will the bag land ?

Solution :

Since, the man and the bag as a system are falling vertically the only external force acting on them is that of gravity which acts along the vertical. There is no horizontal force on this system, thus, if the man throws the bag he applies a force on the bag and according to Newton’s third law the bag will apply an equal and opposite force on the man. So that the man moves towards the pond. But all these forces are internal to the system and thus the c.m. of the system will fall vertically down and will not change its horizontal position.

Let c be the distance travelled by the bag along the horizontal from initial position. To avoid hard ground, the distance travelled by the man will be x. Taking the c.m. at origin.

Thus, bag will fall at a distance

Let the man throw the bag with velocity v towards the direction opposite to the pond.

As the motion of the bag along y-axis is free fall under gravity, equation of motion along y axis.

t = time taken to fall by height (H – h) again if T is the total time to fall by H

Thus, time taken to fall by height h,

Thus, to travel distance along x-axis velocity given by the man = v (Let)

then,

Problem : 6 (IIT. JAM 2009) : A block of mass M has a semicircular cavity as shown below. A particle of mass m is at the point A on the block. The centre of mass of the block is on the dotted line as. Shown at a distance x₁ from the origin. The mass m is now released and it reaches the point B such that the centre of mass of the block is now at distance x₂. Find . (All surfaces are frictionless.)

Solution :

Consider the point mass m and the block of mass together as a system.

Thus, the position of the particle with respect to the origin is initially.

Hence, the initial x coordinate of the centre of mass of the system may be given by

Now, as the particle moves to the position B under the effect of gravity, there’s no external force on the system along x-axis. Thus the position of the c.m. along x-axis should remain intact.

Thus, the coord. of c.m. are :

Hence                

Centre of mass frame of Reference (Brief Discussion) :

Let us go back to our initial problem of a spinning ball. Suppose instead of throwing the ball vertically upwards the spinner throws the ball at an angle w.r.t. the vertical.

Then the motion of any point P on the periphery of the ball is very complicated to visualize but as mentioned earlier the motion of its centre is simpler as it doesn’t undergo rotation. Now we know that if the ball has uniform distribution of mass and spherical symmetry then the centre of mass of the ball coincides with the geometrical centre of the ball.

Thus the motion of the centre of mass can be visualized as a simple projectile motion. Now suppose we attach our frame of reference to the centre of mass of the ball.

Now, in this frame (though non-inertial) the centre of mass of the ball will remain at rest. If we analyze the motion of the point P in this frame then it will be a simple circular motion in a circle of radius (R = Radius of ball).

Thus, the centre of mass frame of reference is a useful tool to deal with complex motions. As will be discussed later in the chapter related to rotation any such complex motion can be broken down into translation of the centre of mass and rotation about the centre of mass.