Understanding Uniform Circular Motion: Exploring Centripetal Acceleration

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Homework Help Overview

The discussion revolves around uniform circular motion, specifically focusing on the nature of centripetal acceleration and its implications for the motion of a particle. Participants explore why a particle in uniform circular motion does not move radially inward despite the presence of radial acceleration.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants examine the relationship between velocity and acceleration in uniform circular motion, questioning how centripetal acceleration affects the direction of motion. Some express confusion about the vector addition of acceleration and velocity, while others seek clarity on the implications of net force being directed radially inward.

Discussion Status

The discussion is active, with multiple participants raising questions and clarifying concepts related to the motion. There is a focus on understanding the dynamics of circular motion, and some participants are exploring different interpretations of how forces and accelerations interact in this context.

Contextual Notes

Participants are grappling with the definitions and implications of centripetal acceleration and net force in the context of uniform circular motion. There is an emphasis on the need for precise understanding of vector quantities and their roles in motion.

andyrk
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In uniform circular motion only a radial component of acceleration is present i.e centripetal acceleration. So why doesn't the particle following uniform circular motion go radially inwards due to radial acceleration?
 
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Acceleration is rate of change of velocity.Velocity is a vector.A change could be due to change in magnitude or change in direction.In uniform circular motion ,only the direction of velocity changes .This centripetal acceleration is responsible for change in the direction of velocity of the object .

To be more clear,draw a velocity vector at time t ,then draw a velocity vector at time t+Δt .See where the difference of the two vectors point .

One thing more,centripetal force mv2/r is not an individual force acting on the object,rather it is the net force acting on the object moving in a circle .It is the same thing as ma in F=ma equation .
 
I mean to say that if the net force of the particle is radially inward then why doesn't the particle move radially inwards? Thanks for the previous reply though :)
 
andyrk said:
I
mean to say that
if the net force of the particle is radially inward then why doesn't
the particle move radially inwards? Thanks for the previous reply though
:)
Lets draw a rough sketch.First draw a circle.Then draw a velocity vector(tangential) at time t.Add acceleration vector(radial).Where do you see the resultant velocity? Is it towards the centre or somewhere tangential to the circle ?
 
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Sorry to pinpoint, but how can we add acceleration vector to velocity vector? Isn't vector addition of only like quantities allowed?
 
andyrk said:
Sorry
to pinpoint,
but how can we add acceleration vector to velocity vector? Isn't vector
addition of only like quantities allowed?

Sorry...I should have been more precise .By adding acceleration ,i meant acceleration times time which has the same dimensions as that of velocity .

You must have encountered the first equation of motion v=u+at. It is a vector equation.
 
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andyrk said:
Sorry to pinpoint, but how can we add acceleration vector to velocity vector? Isn't vector addition of only like quantities allowed?

What we have is: $$\mathbf{a}(t) = \frac{\mathbf{v_f}(t + \Delta t) - \mathbf{v_i}(t)}{\Delta t} \Rightarrow \mathbf{v_f}(t + \Delta t) = \mathbf{a}(t) \Delta t + \mathbf{v_i}(t)$$

So to get the velocity vector at time t + Δt, take the initial velocity vector at t and do a vector sum with the acceleration vector at t multiplied by some scaling factor (which has the correct dimensions, so as to make the eqn sensible).
 
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Referring to other situations of motion

Maybe I haven't understood something what you're trying to say..Ok..so in the equation v=u+at..'t' is 'Δt', 'u' is velocity of the particle 'Δt' time before the time when its velocity is 'v'..'a' is the constant centripetal acceleration..this equation just gives us the direction of vector v by adding vector u to vector at...mathematically as:
\vec{v}=\vec{u}+\vec{a}Δt..so we just get the direction of \vec{v}...how does that tell us where is the particle going to move now? Would it be that particle's direction of motion would be the same as the direction of \vec{v} i.e tangentially? Then it would try to move tangentially at every point but will not as the velocity keeps changing direction? But this is the same as saying that if we throw a ball upwards it would keep going upwards because of the direction of its velocity which is not the case as particle eventually goes in the direction of the net acceleration which is towards the earth..so even if the velocity as you are saying is tangential still the net acceleration is radially inward and so is the net force...So what's with this circular motion case?
 
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andyrk said:
I mean to say that if the net force of the particle is radially inward then why doesn't the particle move radially inwards? Thanks for the previous reply though :)

Imagine for a second that a particle could continue tangentially. Clearly its radial distance would increase infinitely. But that does not happen because at every instant it gets pulled radially inward.
 
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Ok..Thanks!
 
  • #11
andyrk said:
In uniform circular motion only a radial component of acceleration is present i.e centripetal acceleration. So why doesn't the particle following uniform circular motion go radially inwards due to radial acceleration?
By definition, if the particle is uniform circular motion, then it's combination of speed and acceleration result in a circular path.

With only centripetal acceleration, you can create just about any path, circle, ellipse, parabola, hyperbola, sine wave, ... . For example the path a car takes on a twisty road while traveling at constant speed.
 

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