Question

write 3 equations of motion how will you explain uniform circular motion​

Answer 1

Answer:

In physics, circular motion is a movement of an object along the circumference of a circle or rotation along a circular path. It can be uniform, with constant angular rate of rotation and constant speed, or non-uniform with a changing rate of rotation. The rotation around a fixed axis of a three-dimensional body involves circular motion of its parts. The equations of motion describe the movement of the center of mass of a body.

Examples of circular motion include: an artificial satellite orbiting the Earth at a constant height, a ceiling fan's blades rotating around a hub, a stone which is tied to a rope and is being swung in circles, a car turning through a curve in a race track, an electron moving perpendicular to a uniform magnetic field, and a gear turning inside a mechanism.

Since the object's velocity vector is constantly changing direction, the moving object is undergoing acceleration by a centripetal force in the direction of the center of rotation. Without this acceleration, the object would move in a straight line, according to Newton's laws of motion.

Uniform circular motion

In the case of rotation around a fixed axis of a rigid body that is not negligibly small compared to the radius of the path, each particle of the body describes a uniform circular motion with the same angular velocity, but with velocity and acceleration varying with the position with respect to the axis.

The three equations of motion v = u + at; s = ut + (1/2) at2 and v2 = u2 + 2as can be derived with the help of graphs as described below.

1. Derive v = u + at by Graphical Method

Consider the velocity – time graph of a body shown in the below Figure.

Velocity – Time graph to derive the equations of motion.

Velocity-Time graph to derive the equations of motion

The body has an initial velocity u at point A and then its velocity changes at a uniform rate from A to B in time t. In other words, there is a uniform acceleration 'a' from A to B, and after time t its final velocity becomes 'v' which is equal to BC in the graph. The time t is represented by OC. To complete the figure, we draw the perpendicular CB from point C, and draw AD parallel to OC. BE is the perpendicular from point B to OE

Now, Initial velocity of the body, u = OA ------- (1)

And, Final velocity of the body, v = BC -------- (2)

But from the graph BC = BD + DC

Therefore, v = BD + DC -------- (3)

Again DC = OA

So, v = BD + OA

Now, From equation (1), OA = u

So, v = BD + u --------- (4)

We should find out the value of BD now.

We know that the slope of a velocity – time graph is equal to acceleration, a

Thus, Acceleration, a = slope of line AB

or a = BD/AD

But AD = OC = t,

so putting t in place of AD in the above relation, we get:

a = BD/t

or BD = at

Now, putting this value of BD in equation (4) we get

v = at + u

This equation can be rearranged to give:

v = u + at

And this is the first equation of motion.

It has been derived here by the graphical method.

2. Derive s = ut + (1/2) at2 by Graphical Method

Velocity-Time graph to derive the second equation of motion

Velocity–Time graph to derive the equations of motion.

Suppose the body travels a distance s in time t. In the above Figure, the distance travelled by the body is given by the area of the space between the velocity – time graph AB and the time axis OC,which is equal to the area of the figure OABC. Thus

Distance travelled = Area of figure OABC

= Area of rectangle OADC + Area of triangle ABD

We will now find out the area of the rectangle OADC and the area of the triangle ABD.

(i) Area of rectangle OADC = OA × OC

= u × t

= ut ...... (5)

(ii) Area of triangle ABD = (1/2) × Area of rectangle AEBD

= (1/2) × AD × BD

= (1/2) × t × at (because AD = t and BD = at)

= (1/2) at2 ------ (6)

So, Distance travelled, s = Area of rectangle OADC + Area of triangle ABD

or s = ut + (1/2) at2

This is the second equation of motion. It has been derived here by the graphical method.

3. Derive v2 = u2 + 2as by Graphical Method

Velocity-Time graph to derive the third equation of motion

Velocity–Time graph to derive the equations of motion.

We have just seen that the distance travelled s by a body in time t is given by the area of the figure OABC which is a trapezium.

In other words,

Distance travelled, s = Area of trapezium OABC

Distance travelled =Area of trapezium

Now, OA + CB = u + v and OC = t.

Putting these values in the above relation, we get

------- (7)

We now want to eliminate t from the above equation.

This can be done by obtaining the value of t from the first equation of motion.

Thus, v = u + at (First equation of motion)

And, at = v – u or  

Now, putting this value of t in equation (7) above, we get:  

or 2as = v2 – u2 [because (v + u) × (v – u) = v2 – u2]

or v2 = u2 + 2as

This is the third equation of motion.

Answer 2

$\huge\blue{ᴀɴՖᴡᴇʀ}$

Uniform Circulɑr Motion÷

ᴛʜᴇ ᴍᴏᴠᴇᴍᴇɴᴛ ᴏғ ᴀ ʙᴏᴅʏ ғᴏʟʟᴏᴡɪɴɢ ᴀ ᴄɪʀᴄᴜʟᴀʀ ᴘᴀᴛʜ ɪs ᴄᴀʟʟᴇᴅ ᴀ ᴄɪʀᴄᴜʟᴀʀ ᴍᴏᴛɪᴏɴ. ɴᴏᴡ, ᴛʜᴇ ᴍᴏᴛɪᴏɴ ᴏғ ᴀ ʙᴏᴅʏ ᴍᴏᴠɪɴɢ ᴡɪᴛʜ ᴄᴏɴsᴛᴀɴᴛ sᴘᴇᴇᴅ ᴀʟᴏɴɢ ᴀ ᴄɪʀᴄᴜʟᴀʀ ᴘᴀᴛʜ ɪs ᴄᴀʟʟᴇᴅ ᴜɴɪғᴏʀᴍ ᴄɪʀᴄᴜʟᴀʀ ᴍᴏᴛɪᴏɴ. ʜᴇʀᴇ, ᴛʜᴇ sᴘᴇᴇᴅ ɪs ᴄᴏɴsᴛᴀɴᴛ ʙᴜᴛ ᴛʜᴇ ᴠᴇʟᴏᴄɪᴛʏ ᴄʜᴀɴɢᴇs.

$\huge\red{ar = v2r = ω2r}$

$\huge\green{F = ma}$

$\huge\pink{mv2r= mω2r}$