How Do You Calculate Changes in Particle Speed in Circular Motion?

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SUMMARY

The discussion focuses on calculating changes in particle speed during circular motion, specifically using the equations r(t)=3.15 and Θ(t)=8.86t. Key calculations include determining the x-coordinate, y-component of velocity, and magnitude of acceleration at specified times. The participant successfully calculated the x-coordinate (2.52 m), y-component of velocity (22.34 m/s), and acceleration (247 m/s²), but initially struggled with the speed change between t=10 s and t=51 s. Ultimately, it was established that the speed does not change due to the nature of circular motion, where acceleration is perpendicular to velocity.

PREREQUISITES
  • Understanding of polar coordinates and their applications in circular motion
  • Knowledge of differentiation and its application to velocity and acceleration
  • Familiarity with vector components in physics
  • Concept of centripetal acceleration and its relationship with speed
NEXT STEPS
  • Learn how to derive velocity and acceleration from polar coordinates in circular motion
  • Study the concept of centripetal acceleration and its implications on speed
  • Explore the relationship between tangential and radial acceleration in circular motion
  • Practice problems involving the calculation of speed and acceleration in circular paths
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Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators seeking to clarify concepts of velocity and acceleration in circular dynamics.

agargento
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Homework Statement


The motion of a particle moving in a circle in the x-y plane is described by the equations: r(t)=3.15, Θ(t)=8.86t
Where Θ is the polar angle measured counter-clockwise from the + x-axis in radians, and r is the distance from the origin in m.

1. Calculate the x-coordinate of the particle at the time 2.20 s.
2. Calculate the y-component of the velocity at the time 2.20 s?
3. Calculate the magnitude of the acceleration of the particle at the time 1.60 s?
4. By how much does the speed of the particle change from t=10 s to t=51 s?
5. Calculate the x-component of the acceleration at the time 2.60s?

Homework Equations



x(t) = Rcos(ωt)
y(t) = Rsin(ωt)

And their derivatives for velocity and acceleration.

The Attempt at a Solution


[/B]
I got every question except for 4.

1. 2.52 m
2. 22.34 m/s
3. 247 m/s^2
4.
5. 124.13 m/s^2

For 4 I tried a couple of things. The first was to multiply 247*41 but that didn't work. The next thing I did was to use the equation for velocity (The derivatives of the equations above) to find the X and Y component of the velocity of each time (x,y when t=51 and x,y t=10), calculate their magnitude, and then subtract them from each other, didn't work either.

What am I missing?
 
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What's the speed of the particle at t = 10? (How would you calculate it in general?)
 
Doc Al said:
What's the speed of the particle at t = 10? (How would you calculate it in general?)

Differentiate these two

x(t) = Rcos(ωt)
y(t) = Rsin(ωt)

Get the speed for t=10

And then calculate the magnitude of |x'(t), y'(t)|

Or are you asking to demonstrate it with numbers?
 
agargento said:
Differentiate these two

x(t) = Rcos(ωt)
y(t) = Rsin(ωt)
That will work.

So what do you get when you do the differentiation?

And once you have the velocity components, you can find the magnitude of the resultant velocity, which is the speed.
 
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Doc Al said:
That will work.

So what do you get when you do the differentiation?

And once you have the velocity components, you can find the magnitude of the resultant velocity, which is the speed.

x'(t) = -Rsin(ωt)*ω
y'(t) = Rcos(ωt)*ω

I know it should work but for some reason I can't get it to work...

Edit: did a more accurate calculation, the change in speed is 0 m/s :D

But now I'm a bit confused... I found the particle's acceleration... So how does the speed not change? Is it be because it has a constant radial speed, and the acceleration I found is the tangent acceleration? If so, I find it kind of confusing... Can anyone explain how can I know if I need to look at tangent acceleration or radial acceleration...?
 
Last edited:
agargento said:
x'(t) = -Rsin(ωt)*ω
y'(t) = Rcos(ωt)*ω

I know it should work but for some reason I can't get it to work...

Edit: did a more accurate calculation, the change in speed is 0 m/s :D

But now I'm a bit confused... I found the particle's acceleration... So how does the speed not change? Is it be because it has a constant radial speed, and the acceleration I found is the tangent acceleration? If so, I find it kind of confusing... Can anyone explain how can I know if I need to look at tangent acceleration or radial acceleration...?

What makes you think the particle is changing its speed? Isn't it possible to move in a circle at constant speed?
 
PeroK said:
What makes you think the particle is changing its speed? Isn't it possible to move in a circle at constant speed?

It is. But I calculated it's acceleration in 3. and 5.
 
agargento said:
It is. But I calculated it's acceleration in 3. and 5.

Can a particle move in a circle without acceleration?
 
agargento said:
x'(t) = -Rsin(ωt)*ω
y'(t) = Rcos(ωt)*ω

I know it should work but for some reason I can't get it to work...
So far, so good. Now calculate the speed at time t. (You have the components of the velocity.)

agargento said:
But now I'm a bit confused... I found the particle's acceleration... So how does the speed not change? Is it be because it has a constant radial speed, and the acceleration I found is the tangent acceleration? If so, I find it kind of confusing... Can anyone explain how can I know if I need to look at tangent acceleration or radial acceleration...?
Since acceleration is a change in velocity, not necessarily speed, just going in a circle implies some acceleration.
 
  • #10
Doc Al said:
So far, so good. Now calculate the speed at time t. (You have the components of the velocity.)Since acceleration is a change in velocity, not necessarily speed, just going in a circle implies some acceleration.

Ohhh... So velocity has a direction and a quantity (Basically a vector) while speed is just a scalar, so when moving in a circle the velocity always changes, and that's the acceleration... ?
 
  • #11
agargento said:
Ohhh... So velocity has a direction and a quantity (Basically a vector) while speed is just a scalar, so when moving in a circle the velocity always changes, and that's the acceleration... ?

Not just a circle. Any change of direction implies acceleration.

Also, if the force/acceleration is perpendicular the the velocity then the speed does not change. That is what is special about motion in a circle at constant speed: the force/acceleration is always perpendicular to the velocity.
 
  • #12
PeroK said:
Not just a circle. Any change of direction implies acceleration.

Also, if the force/acceleration is perpendicular the the velocity then the speed does not change. That is what is special about motion in a circle at constant speed: the force/acceleration is always perpendicular to the velocity.

Now I understand, thanks!
 

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