Circular Motion: Solving for Velocity using Vector Cross Product

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Homework Help Overview

The discussion revolves around a problem related to circular motion, specifically focusing on calculating velocity using the vector cross product. Participants are examining the relationships between angular velocity, position vectors, and the application of trigonometric functions in a rotational context.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are exploring the calculation of angular velocity and its vector representation. There is discussion about the use of trigonometry to find components of position vectors and the implications of using the right-hand rule for determining vector directions.

Discussion Status

Some participants have provided guidance on the correct interpretation of angular velocity and its vector form. There is an ongoing exploration of how to apply the vector cross product in this context, with various interpretations of the problem being discussed. Participants are actively questioning their assumptions and the clarity of the problem statement.

Contextual Notes

There are mentions of incomplete information in the problem statement, particularly regarding the full figure needed for accurate calculations. Participants are also addressing potential misunderstandings about the relationship between revolutions per second and angular velocity.

connorc234
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Homework Statement


http://i.imgur.com/Nbg3Nrc.jpg

pic of question

Homework Equations


v = wr, s=rtheta t=1/f etc

The Attempt at a Solution


i did a, got 504 radpersec and then did b, r = 1.3sin40 i - 1.3cos40 j + 0 k

but I'm not sure if I'm doing it right. can anyone help me and point me in the right direction?

thanks,
Connor
 
Last edited:
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connorc234 said:

Homework Statement


http://i.imgur.com/oIKdE9U.jpg

pic of question

Homework Equations


v = wr, s=rtheta t=1/f etc

The Attempt at a Solution


i did a, got 504 radpersec and then did b, r = 1.3sin40 i - 1.3cos40 j + 0 k

but I'm not sure if I'm doing it right. can anyone help me and point me in the right direction?

thanks,
Connor
Welcome to the PF.

The top of your image is cut off, so we don't see the whole figure.

Also, could you show the steps that you used to get to your answers? That will save a lot of time in helping you out. :smile:
 
berkeman said:
Welcome to the PF.

The top of your image is cut off, so we don't see the whole figure.

Also, could you show the steps that you used to get to your answers? That will save a lot of time in helping you out. :smile:

sorry, changed it to a better picture.

for a. T = 1/1.4 where T is the period

t = (40/360)(1/1.4) to find the time taken for the mass to turn through 40 degrees
=(1/12.6)

w = dtheta/dt = 40/(1/12.6)
w = (40)(12.6) = 504 radpersec

b. used trigonometry with theta = 40 and hypotenuse = 1.3 to find x and y with the axel centre being the origin

hence 1.3sin40 for x and (-)1.3cos40 for y
 
Your answer for (b) looks right, but (a) doesn't look right.

There are 2π radians in a circle, and the rod rotates through a full circle every 1.4 seconds. What ω does that result in?
 
berkeman said:
Your answer for (b) looks right, but (a) doesn't look right.

There are 2π radians in a circle, and the rod rotates through a full circle every 1.4 seconds. What ω does that result in?

2π/1.4

something i don't understand is that the question is asking for vectors, so how would i put ω = 2π/1.4 in vector form?

should I use my answer for b in the future questions, since it's continuously asking for vectors?
 
connorc234 said:
something i don't understand is that the question is asking for vectors, so how would i put ω = 2π/1.4 in vector form?

should I use my answer for b in the future questions, since it's continuously asking for vectors?
A vector definition of ω involves the cross product:

https://en.wikipedia.org/wiki/Angular_velocity

And the direction of the vector ω is determined by the right hand rule for that cross product. Does that make sense? Have you learned about the vector cross product yet?
 
berkeman said:
A vector definition of ω involves the cross product:

https://en.wikipedia.org/wiki/Angular_velocity

And the direction of the vector ω is determined by the right hand rule for that cross product. Does that make sense? Have you learned about the vector cross product yet?

i've learned the right hand rule yeah, don't think I've seen the cross product before. but i don't know if they just want 2pi/1.4 or if they want me to split it into i's j's and k's somehow. and if so i have no idea how to do that. the fact that they ask for a position vector after that question confuses me too if they do indeed want a vector for the angular velocity
 
connorc234 said:
i've learned the right hand rule yeah, don't think I've seen the cross product before. but i don't know if they just want 2pi/1.4 or if they want me to split it into i's j's and k's somehow. and if so i have no idea how to do that. the fact that they ask for a position vector after that question confuses me too if they do indeed want a vector for the angular velocity
From the right hand rule, since ω ~ r X v, the direction of ω is the direction your right thumb points when you point your right fingers in the direction of r and curl them towards v. What is that vector direction? :smile:
 
berkeman said:
From the right hand rule, since ω ~ r X v, the direction of ω is the direction your right thumb points when you point your right fingers in the direction of r and curl them towards v. What is that vector direction? :smile:
that's the k or z direction right?

so the vector would be 0i + 0j + 2pi/1.4 k ?
 
  • #10
berkeman said:
There are 2π radians in a circle, and the rod rotates through a full circle every 1.4 seconds. What ω does that result in?
If I can just sneak in for a moment,... the problem statement says that there are 1.4 revolutions per second, not 1 revolution per 1.4 seconds. I thought I'd point that out before all the calculations were done :smile:
 
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  • #11
gneill said:
If I can just sneak in for a moment,... the problem statement says that there are 1.4 revolutions per second, not 1 revolution per 1.4 seconds. I thought I'd point that out before all the calculations were done :smile:
Oops, thanks! :smile:
 
  • #12
connorc234 said:
that's the k or z direction right?

so the vector would be 0i + 0j + 2pi/1.4 k ?
Correct, after you apply gneill's fix to the value for ω :smile:
 
  • #13
vbx
berkeman said:
Correct, after you apply gneill's fix to the value for ω :smile:

problem now though...

using the 2 answers for a and b with the equation v=wr in vector form, gives me 0i + 0j + 0k...? is this meant to happen?
 
  • #14
connorc234 said:
vbxproblem now though...

using the 2 answers for a and b with the equation v=wr in vector form, gives me 0i + 0j + 0k...? is this meant to happen?
No. You are going to need to use the vector cross product, or a simplified version if your textbook has it. As shown at this textbook page:

https://books.google.com/books?id=g...QIVT8pjCh2Owgcq#v=onepage&q=v = w x r&f=false

v = r X ω

So you get the vector v using the right hand rule with r and ω. With the problem as drawn, the velocity v is always perpendicular to r, so that simplifies the cross product. What does your textbook list as relevant vector equations for problems like this? You only listed scalar equations...
 
  • #15
berkeman said:
No. You are going to need to use the vector cross product, or a simplified version if your textbook has it. As shown at this textbook page:

https://books.google.com/books?id=gJA2oahuPSMC&pg=PA48&lpg=PA48&dq=v+=+w+x+r&source=bl&ots=SRNGSMXm_F&sig=aR2L-m4L_YkUZJYRmKyL7gBA01o&hl=en&sa=X&ved=0CFkQ6AEwCGoVChMIjq-LxOOByQIVT8pjCh2Owgcq#v=onepage&q=v = w x r&f=false

v = r X ω

So you get the vector v using the right hand rule with r and ω. With the problem as drawn, the velocity v is always perpendicular to r, so that simplifies the cross product. What does your textbook list as relevant vector equations for problems like this? You only listed scalar equations...

http://i.imgur.com/VL6r6Oo.jpg

i have this which i think might be relevant. does this mean i can just use v=wr as normal?
 
  • #16
connorc234 said:
http://i.imgur.com/VL6r6Oo.jpg

i have this which i think might be relevant. does this mean i can just use v=wr as normal?
Yes, and the direction is the vector shown in the diagram. You can figure out the x and y components given the angles...
 
  • #17
berkeman said:
Yes, and the direction is the vector shown in the diagram. You can figure out the x and y components given the angles...
ok thanks.

one last question. should i use the position vector for r (1.3sin40 i -1.3cos40 j) or its modulus?
 
  • #18
Depends. If you were doing the vector cross product, then you would use the components. Since you are using the simplified form, you need the magnitude of r and the magnitude of ω multiplied together. You get the magnitude of v with that multiplication, and then you convert it into the vector components of v. Makes sense?
 
  • #19
berkeman said:
Depends. If you were doing the vector cross product, then you would use the components. Since you are using the simplified form, you need the magnitude of r and the magnitude of ω multiplied together. You get the magnitude of v with that multiplication, and then you convert it into the vector components of v. Makes sense?

yeha i got it now thanks a lot man really appreciate it.
 
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