How Do You Calculate Jerk in Uniform Circular Motion?

  • Thread starter toesockshoe
  • Start date
  • Tags
    Jerk Stuck
In summary, the question asks to calculate the jerk of an object in uniform circular motion, both in terms of direction and magnitude. The direction of the jerk is opposite the velocity vector, while the parallel component is 0 due to the parallel component of acceleration being 0 in UCM. To find the magnitude of the perpendicular component, one can use the equations v=ωr and a=αr, where ω is the angular velocity and r is the radius of the circle. From there, the position, velocity, acceleration, and jerk vectors can be compared to determine their magnitudes and directions.
  • #1
toesockshoe
265
2

Homework Statement


The question asks to calculate the jerk of an object (both its direction and magnitude) in uniform circular motion.

Homework Equations


j=d(a)/dt

The Attempt at a Solution


I know that the direction is opposite the velocity vector (I did this by drawing a circle and taking a limit of average acceleration as t approaches 0). I also know that the parallel component of jerk is 0 because the parallel component of acceleration in UCM is 0. I am stuck on the magnitude of the perpendicular component. I know that
[tex] \vec{a}_{\perp} = \frac{v^2}{2}(-r) [/tex] (the r is a unit vector)... can I just take the derivative of this? wouldn't that be 0? because v^2/r is a constant and there is no t. i am suck.
 
Last edited:
Physics news on Phys.org
  • #2
Perhaps start with a parameterized version of the motion?

##r(t) = R (cos(\omega t) \vec{i} + sin(\omega t) \vec{j})##

Should be easy enough to differentiate repeatedly...
 
  • #3
gneill said:
Perhaps start with a parameterized version of the motion?

##r(t) = R (cos(\omega t) \vec{i} + sin(\omega t) \vec{j})##

Should be easy enough to differentiate repeatedly...
where did you get that parameterized version?
 
  • #4
It's just a conversion from polar form to rectangular form of a circle. x = R cos(θ), y = R sin(θ), where θ = ωt to make it time dependent.
 
  • #5
gneill said:
It's just a conversion from polar form to rectangular form of a circle. x = R cos(θ), y = R sin(θ), where θ = ωt to make it time dependent.
ok so, j_x = Rw^3sin(wt) and j_y = -Rw^3cos(wt) right?

the problem is however that my prof wanted us to express it interms of v and r. i don't really know how to convert from w to r... we haven't really talked about w yet.
 
  • #6
It seems strange that you'd be learning about a concept like jerk without having covered the basics of rotational motion.

##v = \omega r~~ ; ~~a = \alpha r## are the basic relationships between angular and linear velocities and accelerations. You would profit from taking the magnitudes of each of the vectors along the differentiation path: position → velocity → acceleration → jerk. For example, the magnitude of the velocity vector is v = ω R.
 
  • #7
gneill said:
It seems strange that you'd be learning about a concept like jerk without having covered the basics of rotational motion.

##v = \omega r~~ ; ~~a = \alpha r## are the basic relationships between angular and linear velocities and accelerations. You would profit from taking the magnitudes of each of the vectors along the differentiation path: position → velocity → acceleration → jerk. For example, the magnitude of the velocity vector is v = ω R.
yeah, it was just one problem out of a list of 10 or so. our teacher kinda talked about w in the last 5 minutes of class on friday, but we haven't touched rotations yet... we haven't even begun f=ma (school just started)
 
  • #8
toesockshoe said:
the problem is however that my prof wanted us to express it interms of v and r. i don't really know how to convert from w to r... we haven't really talked about w yet.

toesockshoe said:
our teacher kinda talked about w in the last 5 minutes of class on friday,

Did your teacher mention that v2/r = rw2?

Most classes have reading assignments in addition to classroom lectures, so maybe it's discussed in more depth there.
 
  • #9
Mister T said:
Did your teacher mention that v2/r = rw2?

Most classes have reading assignments in addition to classroom lectures, so maybe it's discussed in more depth there.
he wrote on the board that v=ds/dt = r d(theta)/dt ... and d(theta)/dt is the same as w. that is all he talked/wrote about w. so sorry Mister T... it was hidden in my notes... i guess he did write an equation relating the 2.
 
  • #10
toesockshoe said:
ok so, j_x = Rw^3sin(wt) and j_y = -Rw^3cos(wt) right?

Right. So looking at the x-components and the y-components of the position vector and the acceleration vector, what can you conclude about the direction of those two vectors?

Now make the same comparison of the velocity vector and the jerk vector.

Can you find the magnitude of these vectors from their components?

These should lead you to being able to describe the magnitude and direction of the jerk vector, which is what your teacher is asking you to do.
 

Related to How Do You Calculate Jerk in Uniform Circular Motion?

1. What is UCM?

UCM stands for Uniform Circular Motion. It refers to the motion of an object traveling in a circular path at a constant speed.

2. What is jerk in UCM?

Jerk is the rate of change of acceleration in UCM. It describes how quickly the acceleration of an object changes as it moves along a circular path.

3. Why is it important to calculate jerk in UCM?

Calculating jerk in UCM can help us understand the dynamics of an object's motion and how it is affected by external forces. It also allows us to make predictions about the object's future motion.

4. What are the units for jerk in UCM?

The units for jerk in UCM are meters per second cubed (m/s^3) or radians per second cubed (rad/s^3).

5. How do you calculate jerk in UCM?

Jerk in UCM can be calculated by finding the third derivative of the position function with respect to time, or by dividing the change in acceleration by the change in time.

Similar threads

  • Introductory Physics Homework Help
2
Replies
55
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
543
Replies
6
Views
279
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
303
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
2K
Back
Top