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Uniform circular motion

The object moves along a circular path constant speed.

Notwithstanding the constant magnitude, acceleration of uniform circular motion is a variable acceleration in the horizontal plane, because direction of radial centripetal acceleration (shown with red arrow) keeps changing with time. Therefore, the acceleration of the motion keeps changing and is not independent of time as required for acceleration to be constant.

Equation of motion

The motion under constant acceleration allows us to describe accelerated motion, using simple mathematical construct. Here, we set out to arrive at these relationships in the form of equations. In these equations, “ u ” stands for initial velocity, “ v ” for final velocity, “ a ” for constant acceleration and “t” for time interval of motion under consideration. This is short of convention followed by many text books and hence the convention has been retained here.

First equation

1: v = u + a t

This relation can be established in many ways. One of the fundamental ways is to think that velocity is changed by constant acceleration (vector) at the end of successive seconds. Following this logic, the velocity at the end of t successive seconds are as under :

At the end of 1 st second : u + a At the end of 2 nd second : u + 2 a At the end of 3 rd second : u + 3 a At the end of 4 th second : u + 4 a ------------------------------- ------------------------------- At the end of t th second : u + t a = u + a t

We can also derive this equation, using the defining concept of constant acceleration. We know that :

a = a avg a = Δ v Δ t = v - u t a t = v - u v = u + a t

Alternatively (using calculus), we know that :

a = đ v đ t đ v = a đ t

Integrating on both sides, we have :

Δ v = a Δ t v 2 - v 1 = a t v - u = a t v = u + a t

Second equation

2: v avg = u + v 2

The equation v = u + a t is a linear relationship. Hence, average velocity is arithmatic mean of initial and final velocities :

Average velocity

Average velocity is equal to mean velocity.

v avg = u + v 2

Third equation

3: s = Δ r = r 2 - r 1 = u t + 1 2 a t 2

This equation is derived by combining the two expressions available for the average velocity.

v avg = Δ r Δ t = r 2 - r 1 t

and

v avg = ( v 1 + v 2 ) 2 = ( v + u ) 2

Combining two expressions of average acceleration and rearranging, we have :

( r 2 - r 1 ) = ( v + u ) t 2

Using the relation, v = u + a t and substituting for v, we have :

( r 2 - r 1 ) = ( u + a t + u ) t 2 s = Δ r = r 2 - r 1 = = u t + 1 2 a t 2

Alternatively, we can derive this equation using calculus. Here,

v = đ r đ t đ r = v đ t

Integrating between the limits on both sides,

đ r = v đ t

Now substituting v ,

đ r = ( u + a t ) đ t Δ r = u t + a t đ t s = Δ r = r 2 - r 1 = u t + 1 2 a t 2

The expression for the displacement has two terms : one varies linearly ( u t) with the time and the other ( 1 2 a t 2 ) varies with the square of time. The first term is equal to the displacement due to non-accelerated motion i.e the displacement when the particle moves with uniform velocity, u . The second term represents the contribution of the acceleration (change in velocity) towards displacement.

This equation is used for determining either displacement(Δ r ) or position( r 1 ). A common simplification, used widely, is to consider beginning of motion as the origin of coordinate system so that

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
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Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
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David Reply
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David
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emma Reply
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Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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Adjanou
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Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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Maurice Reply
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answer
Magreth
progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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