IEMDaily - Video Lecture Notes

Definition

Center of mass of continuous three dimensional objects

Center of mass of continuous three dimensional objects is found by:

{\vec{r}}_{C O M} = \frac{\int \vec{r} d m}{\int d m}

Example:
Distance of the center of mass of a solid uniform cone from its vertex is

z_{0}

. If the radius of its base is R and its height is h then find

z_{0}

.
Solution:
Suppose the cylindrical symmetry of the problem to note that the center of mass must lie along the

z

-axis (

x = y = 0

). The only issue is how high does it lie. If the uniform density of the cone is

ρ

, then first compute the mass of the cone. If we slice the cone into circular disks of area

π r^{2}

and height

d z

, the mass is given by the integral:

M = \int ρ d V = ρ \int_{0}^{h} π r^{2} d z

However, we know that the radius

r

starts at a for

z = 0

, and goes linearly to zero when

z = h

. This means that

r = a (1 - z / h)

, so that:

M = ρ \int_{0}^{h} π a^{2} (1 - z / h)^{2} d z = π a^{2} ρ \int_{0}^{h} (1 - 2 z / h + z^{2} / h^{2}) d z = 1 / 3 π a^{2} h ρ

Now this simply indicates that the volume of the cone is given by

V = \frac{1}{3} π a^{2} h

To find the height of the center of mass, we then compute:

z_{c m} = \frac{1}{M} \int ρ z d V = \frac{ρ}{M} \int_{0}^{h} π r^{2} z d z = \frac{π a^{2} ρ}{M} \int_{0}^{h} (1 - \frac{z}{h})^{2} z d z

= \frac{π a^{2} ρ}{M} \int_{0}^{h} (z - \frac{2 z^{2}}{h} + \frac{z^{3}}{h}) d z = \frac{1}{12 M} π a^{2} h^{2} ρ = \frac{1}{4} h

Note:
Choice of correct coordinate system and correct elemental sections is very important to make calculations easier.

Example

Centre of Mass of continuous one dimensional bodies

Example: A non-uniform rod having mass per unit length as

μ = a x

(a is constant). If its total mass is M and length L.Find position of the centre of mass.

Solution:
Choose a coordinate system with the rod aligned along the x-axis and origin located at the left end of the rod. Choose an infinitesimal mass
element dm located a distance x'. Let the length of the mass element be dx'.
Thus

d m = μ (x^{'}) d x^{'}

The total mass is found by integrating the mass element over the length of the rod

M = \int_{0}^{L} μ (x^{'}) d x^{'} = a \int_{0}^{L} x^{'} d x^{'} = \frac{a}{2} x^{' 2} ∣_{0}^{L} = \frac{a}{2} L^{2}

a = \frac{2 M}{L^{2}}

Now center of mass is calculated as

x_{c m} = \frac{1}{M} \int_{b o d y} x d m = \frac{1}{M} \int_{0}^{L} μ (x^{'}) x^{'} d x^{'} = \frac{a}{M} \int_{0}^{L} x^{' 2} d x^{'}

substituting the value of a

\frac{2}{L^{2}} \int_{0}^{L} x^{' 2} d x^{'} = \frac{2}{3 L^{2}} x^{' 3} ∣_{0}^{L} = \frac{2}{3 L^{2}} (L^{3} - 0) = \frac{2}{3} L

Definition

Centre of mass of Continuous Bodies

Centre of mass of bodies coincide with their geometric centres and this this can be determined by method of symmetry. For instance centre of mass of a uniform semicircular disc lies on the vertical axis as the object is symmetric about this.

Formula for finding centre of mass of continuous system:

X_{c m} = \frac{\int r d m}{\int d m}

Definition

Velocity of Centre of Mass

V_{c m} = \frac{m_{1} v_{1} + m_{2} v_{2} + . . .}{m_{1} + m_{2} + . . .}

The velocity of centre of mass for a system of particles is defined this way and given by above formula.

Example

Acceleration of Centre of Mass

Acceleration of centre of mass is defined by:

a_{c m} = \frac{m_{1} a_{1} + m_{2} a_{2} + . . . m_{n} a_{n}}{m_{1} + m_{2} + . . . m_{n}}

Example

Example on change in position of centre of mass

Example: A mass m is at rest on a inclined plane structure of mass M which is further resting on a smooth horizontal plane. Now if the mass starts moving, Find how the position of centre of mass of the system will change.

Solution:
Here the system is wedge+block. Net force on the system in horizontal direction is "0". Hence the centre of mass of the system will not move in horizontal direction. Now for vertical direction there is a force that is due to the mass of the wedge and the block and hence the centre of mass changes in this direction.

Example

Position of centre of mass of two blocks attached in spring

Example: An elastic spring is compressed between two blocks of masses

1 k g

and

2 k g

resting on a smooth horizontal table as shown. If the spring has

12 J

of energy and suddenly released, the velocity with which the larger block of

2 k g

moves will be:

Solution:Using momentum conervation

m_{A} . v_{A} = m_{B} v_{B}

\Rightarrow v_{A} = 2 v_{B}

Now total K.E=P.E of spring

\frac{1}{2} m_{A} v_{A}^{2} + \frac{1}{2} m_{B} {(v_{B})}^{2} = 12

\frac{1}{2} \frac{m_{B}}{2} \times 4 v_{B}^{2} + \frac{1}{2} m_{B} (v_{B})^{2} = 12

3 {(v_{B})}^{2} = 12

v_{B} = 2 m / s

Definition

Centre of Gravity

The centre of gravity (C.G.) of a body is the point about which the algebraic sum of moments of weights of all the particles constituting the body is zero. The entire weight of the body can be considered to act at this point howsoever the body is placed.

Definition

Centre of gravity

Centre of gravity is a point from which the weight of a body or system may be considered to act. Whereas, the center of mass is the point where all of the mass of the object is concentrated. In uniform gravity it is the same as the centre of mass.

Definition

Shift of center of gravity and its effect

The location of the center of gravity is important for stability. If we draw a line
straight down from the center of gravity of an object of any shape and it falls inside the base of the object, then the object will be stable.
If the line through the center of gravity falls outside the base then the object will be unstable.
For example:When you stand erect the center of gravity is somewhere near to your stomach and your body is balanced but when you try to lean forward or backward the center of gravity shifts outside the base the and your body gets unstable.