Mechanics | Physics and Universe https://physicsanduniverse.com Fri, 03 Oct 2014 06:46:18 +0000 en-US hourly 1 https://wordpress.org/?v=6.8.6 52870843 Simple Pendulum https://physicsanduniverse.com/simple-pendulum/ https://physicsanduniverse.com/simple-pendulum/#respond Fri, 03 Oct 2014 06:46:18 +0000 http://physicsanduniverse.com/?p=5095 A simple pendulum is a heavy mass (bob) suspended by very low weight inextensible and flexible string which undergo to and fro motion in a plane from mean position. This to and fro motion of pendulum becomes simple harmonic motion for small angle of oscillation. Let is consider a simple pendulum with mass ‘m’ and length of the string as ‘l’. Let us displace the bob from mean position by small angle \theta . This displacement should be as small as possible. Once this is displaced we release the bob and we can see the oscillation from mean position. Let us breakdown the force acting on this bob simple-pendulum

1. mg, the weight of bob acting vertically downward.

2. Tension of the string T with direction towards the point of suspension

3. mg sin\theta acting towards the mean position

4. mg cos\theta acting away from suspension point

When the bob is at rest the component T and mg cos\theta act opposite to each other and these forces balances out each other.

The force mg sin\theta tries to bring the bob towards the mean position  and this force is called restoring force.  During the process or restoration to mean position, an acceleration ‘a’ is produced towards the mean position so,

-mgsin\theta = ma negative sign indicates that force is opposite to displacement.

-g sin\theta = a

\theta is as stated earlier is very small so, in radian system of measurement we can approximate like sin\theta \equiv \theta

a = -g\theta

We also have

\theta = \dfrac{Arc}{Radius} = \dfrac{Arc AB}{l}

If we assume that l is very long and \theta is very small (less than 4 degree) then the arc AB will be almost equal to linear displacement ie AB. So, Arc AB = x

\theta = \dfrac{x}{l}

Using this value of \theta in a = -g\theta we get

a = -g \dfrac{x}{l} = - \dfrac{g}{l} x

\dfrac{a}{x} = - \dfrac{g}{l} ……………. (1)

The negative sign indicate that the acceleration is towards the mean position. In a pendulum we can assume that length of pendulum ‘l’ is constant and so is ‘g’. Hence we get

a \propto x

Now see that the acceleration of bob is proportional to displacement and is directed towards the mean position, we can say the the motion is Simple Harmonic Motion. Therefore we have

\dfrac{a}{x} = -\omega^2 =-(\dfrac{2\pi}{T})^2 ………….. (2)

Comparing equations (1) and (2) we get

(\dfrac{2\pi}{T})^2 = \dfrac{g}{l}

T = 2\pi \sqrt{\dfrac{l}{g}}

This equation gives the time period of a simple pendulum if we know the value of ‘g’ and ‘l’.

]]> https://physicsanduniverse.com/simple-pendulum/feed/ 0 5095 Circular Motion https://physicsanduniverse.com/circular-motion/ https://physicsanduniverse.com/circular-motion/#respond Wed, 24 Sep 2014 06:12:04 +0000 http://physicsanduniverse.com/?p=3364 Circular motion refers to the movement of an object in a circular path. By circular we mean motion about a fixed point. The most suitable examples of circular motion includes motion of the planet around the sun (they are not exactly circular but calculation gets easier and we approximate them to be circular) and stone whirled in circle by attaching it to a string.

Angular Displacement

In circular motion, the distance of object from a fixed point remains the same. The angle described at the center of the circle by the an object moving in the circle is called angular displacement. Its unit is radian. It is defined as the ratio of arc length to radius of circle

\theta = \dfrac{Arc}{radius}

It is a vector quantity with direction perpendicular to the plane of the circle.

Angular Velocity

It is the angle described at the center in one second by a particle moving in a circle. Mathematically, we can write angular velocity as

\omega = \dfrac{\Delta\theta}{\Delta t}

Its unit is rad/s. Dividing above equation by \Delta t we get

\dfrac{\triangle\theta}{\triangle t} = \dfrac{1}{r} \dfrac{\triangle s}{\triangle t}

\omega = \dfrac{v}{r} where v = \dfrac{\triangle s}{\triangle t}  is linear velocity.

This is also the relation connecting linear velocity and angular velocity of an object moving in a circle of radius r. Angular velocity is a vector quantity with direction perpendicular to the plane of circle.

Frequency and time period in circular motion

Frequency is the number of complete revolutions made in one second by an object moving is a circle. Time period is the time required to make one complete rotation in a circle.

Time period T = \frac{1}{f}

Angular velocity

\omega = \dfrac{2\pi}{T} = 2 \pi f

Centripetal acceleration and centripetal force

Any object moving is a circular path of radius r and uniform speed v has an acceleration towards the center and this acceleration is given by

a = \omega^2 r \dfrac{v^2}{r}

To produce this acceleration a force is necessary which is also directed towards the center. This force is centripetal force and is given by

F = ma = \dfrac{mv^2}{r} = m \omega^2 r

This force is directed towards the center.

Centrifugal force

Centrifugal force is the apparent force that draws a rotating body away from the center of rotation. Centrifugal force is known as an outward force visible in a rotating frame of reference. It is also called a fictitious force in the sense that it is not part of an interaction but is a result of rotation.

]]> https://physicsanduniverse.com/circular-motion/feed/ 0 3364 Variation of g https://physicsanduniverse.com/variation-of-g/ https://physicsanduniverse.com/variation-of-g/#comments Mon, 15 Sep 2014 08:57:14 +0000 http://physicsanduniverse.com/?p=1756 The value of g at the surface of the earth is given by the equation g=G\frac{GM}{R^2} and its average value on the surface of the earth is about 9.8 ms^{-2} . But we have to remember that the value of g changes with R and some other factors which we will  discuss below. Even from above equation we can tell that as the body moves above or below the surface of the earth, the value of g changes.

Variation of g above the surface of the Earth

Let us put a body at the height h above the surface of the earth. Now the distance of the object from the center of the earth is (R+h). Therefore the acceleration due to gravity at this point is given by the equation

g_h=\dfrac{GM}{(R+h)^2}

Acceleration due to gravity at the surface of the earth is

g=\dfrac{GM}{R^2}

Dividing these equation we get

\dfrac{g_h}{g} = \dfrac{R^2}{(R+h)^2}

From this we can say that (R+h)^2 will be greater than R^2 and this will lead to a conclusion that g_h < g . Hence acceleration due to gravity decreases as the height increases.

For more approximation lets say that height of body (h) is far far less than radius of earth (R) and using binomial expansion we get

\dfrac{1}{(R+h)^2} = (R+h)^{-2} = \dfrac{1}{R^2} ( 1 - \dfrac{2h}{R} + ... )

Hence we have the final result as

\dfrac{g_h}{g} = ( 1 - \dfrac{2h}{R} )

Variation of g below the surface of the Earth

Let us assume that an object is d depth below the surface of the earth. Then in this condition, the distance of the object from the center of the earth is (R-d). In this situation, the whole mass of the earth will not attract the object. The only mass attracting the object will be the volume of sphere having radius (R-d). If we consider earth as homogenous then the effective mass attracting the object is

M_{eff} = \dfrac{4}{3} \pi (R-d)^3 \rho

Now, acceleration due to gravity below the surface of the earth is

g_d = G \dfrac{M_{eff}}{(R-d)^2} = G \dfrac{\frac{4}{3}\pi (R-d)^3 \rho}{(R-d)^2}

g_d = \dfrac{4}{3} \pi G (R-d) \rho   ….. (1)

Mass of earth is

M_E = \dfrac{4}{3} \pi R^3 \rho

Now, putting the mass of earth in equation g=\frac{GM}{R^2} we get

g = \dfrac{4}{3} \pi G R \rho   ……. (2)

Dividing equation (1) by (2) we get

g_d = g(1 - \dfrac{d}{R})

This expression shows that the value of g decreases as we the depth ‘d’ increases. This shows that as we go deep into the center of the earth, the value of g decreases continuously and at the center of the earth it becomes zero.

So, we find that acceleration due to gravity decreases as we move above or below the surface of the earth. The value of g is maximum at the surface of the earth.

 Variation of g due to rotation of Earth

Due to earth’s rotation, the value of g increased while we move from equator to the poles along the surface of the earth. So, the value of g is maximum at pole and minimum at equation. The relation for this variation is given by

g' = g \sqrt{1 - \dfrac{2 \omega^2 R cos^2 \phi}{g}}

Here \phi is the latitude of the place, R is the radius of the earth and \omega is the angular velocity of earth.

Variation of g due to shape of the Earth

Earth is ellipsoidal and not perfectly spherical. So, it is flatter at pole and has bulge at the equator. The polar radius of earth is 6.357 \times 10^6 m and equatorial radius is 6.378 \times 10^6 m Since acceleration due to gravity is inversely proportional to the radius of the earth, g at pole is greater than that at the equator.

]]> https://physicsanduniverse.com/variation-of-g/feed/ 1 1756 Newton’s law of Gravitation https://physicsanduniverse.com/newtons-law-gravitation/ https://physicsanduniverse.com/newtons-law-gravitation/#respond Mon, 08 Sep 2014 07:28:35 +0000 http://physicsanduniverse.com/?p=757 The motion of heavenly objects like Planets, sun, natural satellites etc has drawn a lot of attention in the past the the mystery of their motion remain unsolved until 1665. Sir Issac Newton at the age of 23 made one of the biggest contribution to Physics by simply showing that the force by which moon revolves around the Earth is same force responsible for causing apple to fall to the ground. Sir Issac Newton extended this idea and mad his famous law of Gravitation. The statement of this law states that

Every body in the Universe attract another body with a force called Gravitational force and this force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Let is assume two bodies with masses m_1 and m_2 kept at a distance of r from the centers of the body. If we represent the gravitational force between these two bodies by F then

F \propto m_1m_2 and

F \propto \frac{1}{r^2}

Combining these two relations we get,

F \propto \frac{m_1m_2}{r^2}

or, F = G \frac{m_1m_2}{r^2}

Here G is a constant called Universal gravitational constant and its value is 6.67 \times 10^{-11} Nm^2kg^{-2}

G can also be defined  as – When to object of masses 1 kg each are kept at a distance of 1m then the gravitational force produced between these two objects is 6.67 \times 10^{-11} N

Gravitational Field  Intensity

Gravitational field intensity at any point on the gravitational field is the force exerted by the body on unit mass placed at that point. The value of gravitational field intensity decreases as we move further from the body. The formula for Gravitational field intensity is

Intensity = G \frac{M_E}{R^2}

The unit of Gravitational field intensity is Nkg^{-1}

Acceleration due to Gravity

The gravity of the earth is the force on a body near the surface of the earth and this force is given by

F=G\frac{M_Em}{R_E^2}

Newton’s second law of motion  says the force produced in a body is product of its mass and acceleration

F=ma

Comparing these two equations we get

a=G\frac{M_E}{R_E^2}

This acceleration is represented by g and is called acceleration due to gravity and is given by

g=G\frac{M_E}{R_E^2}

The standard value of acceleration due to gravity ‘g’ on the surface of the earth is 9.81ms^{-2} . This relation shows that the acceleration due to gravity is independent of the mass of the body. This is the reason all body fall to the surface of the earth at the same time irrespective of their masses. Famous feather and coin experiment demonstrates this fact.

Mass of Earth

We now have g=G\frac{M_E}{R_E^2}
After some manipulation we get the mass of earth as
M_E = g\frac{R_E^2}{G}

Density of Earth

To find the density of earth we make some assumptions. We consider earth as homogenous sphere of radius R_E The volume of this sphere is given by
V = \frac{4}{3}\pi R_E^3
Therefore the density of earth is
\rho_E = \frac{M_E}{V}
\rho_E = \frac{3g}{4\pi GR_E}

]]> https://physicsanduniverse.com/newtons-law-gravitation/feed/ 0 757 Elastic collision in one dimension https://physicsanduniverse.com/elastic-collision-one-dimension/ https://physicsanduniverse.com/elastic-collision-one-dimension/#respond Sun, 20 Jul 2014 15:32:46 +0000 http://physicsanduniverse.com/?p=694 Any collision is elastic if the total kinetic energy of the colliding particles remains conserved. Let us consider two bodies A and B with masses m_1 and m_2 are moving with the initial velocity u_1 and u_2 respectively in the same direction and same straight line. In this problem let us suppose that velocity of one object is greater than other (u_1 > u_2) and they are on the collision path. In this situation object A will collide with B and this is called head on collision. After collision and according to our assumption velocity of A will decrease to v_1 and velocity of B will increase to v_2 If both objects are moving on the same direction after collision then we can say that
Total initial momentum of A and B before collision = m_1u_1 + m_2u_2
Total final momentum of A and B after collision = m_1v_1 + m_2v_2

According to conservation of momentum principle
m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2
m_1(v_1-u_1) = m_2(u_2-v_2) ——- (1)

Total kinetic energy of the particles before collision
KE_i = \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2

Total kinetic energy of the particles after collision
KE_f = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

For perfectly elastic collision KE_i = KE_f
KE_i = \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2
m_1(v_1^2-u_1^2) = m_2(u_2^2-v_2^2)
m_1(v_1-u_1)(v_1+u_1) = m_2(u_2-v_2)(u_2+v_2)
Dividing above equation by (1) we get
(u_1-u_2) = -(v_1 - v_2) ——- (2)

Here (u_1-u_2) is the relative velocity of approach of A towards B and (v_1-v_2) is the relative velocity of separation of B and A.

Equation (2) can also be written as
u_1+v_1 = u_2+v_2

Let us multiply above equation by m_2 and add equation (1) followed by some rearrangement we get
v_1 = \frac{m_1-m_2}{m_1+m_2}u_1 + \frac{2m_2}{m_1+m_2}u_2

Similarly let us multiply above equation by m_1 and subtracting equation (1) followed by some rearrangement we get
v_2 = \frac{2m_1}{m_1+m_2}u_1 + \frac{m_2-m_1}{m_1+m_2}u_2

Note 1: If mass of A is very very higher than B we can see that the velocity of A remains unchanged while that of body B changes after collision.
Note 2: If object A is much more smaller than B then velocity of A is changed but the velocity of B remains same.
Note 3: If mass of both body A and B are equal then the velocity if the particles are interchanged. v_1=u_2 and v_2=u_1

]]> https://physicsanduniverse.com/elastic-collision-one-dimension/feed/ 0 694 Projectile motion in Horizontal direction https://physicsanduniverse.com/projectile-motion-horizontal-direction/ https://physicsanduniverse.com/projectile-motion-horizontal-direction/#respond Mon, 30 Jun 2014 04:53:22 +0000 http://physicsanduniverse.com/?p=683 Let us suppose that an object is thrown from a certain height ‘h’ above the ground in horizontal direction with initial velocity ‘u’. This horizontal velocity remains unaffected by the acceleration due to gravity. In this case ‘g’ remains constant throughout the motion. There are two way motion in the projectile, one is vertical motion and another is horizontal motion.

Vertical motion
Just after the projectile is thrown, its initial vertical velocity is zero. The vertical velocity will increase with time and is give by
v=u+gt
u=0
v=gt …. (1)

Now to calculate the time to reach the ground from the height h we have
h=ut + \frac{1}{2}gt^2
h=\frac{1}{2}gt^2
t=\sqrt{\frac{2h}{g}}

Horizontal motion
The time taken by the projectile to reach the ground is calculated above. During this time the projectile moves in the horizontal direction and covers certain distance. This distance is called range. The horizontal range covered is given by the equation
R=ut+\frac{1}{2}at^2
a=0 in horizontal direction
R=ut
Here t is the time projectile stays in the air and is calculated above. Hence using that equation we can calculate range as
R=u\sqrt{\frac{2h}{g}}

]]> https://physicsanduniverse.com/projectile-motion-horizontal-direction/feed/ 0 683 Escape velocity https://physicsanduniverse.com/escape-velocity/ https://physicsanduniverse.com/escape-velocity/#respond Mon, 14 Apr 2014 17:12:13 +0000 http://physicsanduniverse.com/?p=620 We all know that everything thrown vertically upward will eventually fall down. When we throw an object, it moves up to a maximum height and then comes back to the surface. The maximum height attained depends on the velocity of the object. If we talk about the energy, the Kinetic Energy present in the body is used in doing work against gravity. As the body moves upward, its kinetic energy keeps on decreasing and the corresponding amount of potential energy keeps on increasing. When the object is at the maximum height, all kinetic energy is converted into potential energy and the velocity of object becomes zero. Once the velocity is zero, the body starts to fall under the influence of gravity.

So, if we slowly increase the speed of the object, maximum height attained also keeps on increasing and a stage will come when the body will escape the gravitational field of Earth and will never return to the surface.

The minimum velocity of a body that will just take it outside the gravitational field of earth (or any other planet/moon) is called as the Escape velocity.

Escape velocity of Earth

To find the escape velocity of an object let’s suppose that m is the mass of the body and v is the initial velocity from the surface of the earth. Now the kinetic energy of the body is given by,

K.E. = \frac{1}{2}mv^2

The potential energy of the body at the surface of earth is

P.E. = -G\frac{M_em}{R_e}

Total energy of the body at the surface of earth is

K.E. + P.E. = \frac{1}{2}mv^2 - G\frac{M_em}{R_e}

Let assume that initial kinetic energy is just good enough to carry object out of the gravitational field. In this situation, the kinetic energy of the body becomes zero when it escapes the gravitational field and all kinetic energy is used up in doing work against gravity and as explained above gets converted into potential energy. Since gravitational potential energy is negative, the maximum P.E. of a body in the earth’s gravitational field in zero which is at infinity. So, the total energy of the body just outside the earth’s gravitational field is zero.

K.E. + P.E. = 0

Earth’s gravitational field is conservative force field so the total energy remains constant at any point in the field.

\frac{1}{2}mv^2 - G\frac{M_em}{R_e} = 0

v^2 = \frac{2GM_e}{R_e}

v^2 = 2gR_e

since g = \frac{GM_e}{R_e^2}

so, v_{escape} = \sqrt{2gR_e}

To calculate the escape velocity of earth, we just have to plug in some numbers like g = 9.8 m/s^2 and mean radius of earth is R_e = 6.368 \times 10^6 m

v_{escape} = \sqrt{2 \times 9.8 \times 6.368 \times 10^6} = 11.17 \times 10^3 m/s

This implies that if a body is thrown with a velocity of 11.2 km/s from the surface of the earth, it will never come back to the surface of the earth.

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