Circular Motion: Solving for Velocity using Vector Cross Product

AI Thread Summary
The discussion focuses on solving a circular motion problem involving angular velocity and vector cross products. The user initially calculated angular velocity incorrectly but was corrected to use the formula based on the number of revolutions per second. They learned that the vector form of angular velocity can be represented using the right-hand rule and the cross product. The conversation also clarified the use of position vectors and magnitudes in calculating velocity. Ultimately, the user gained a better understanding of how to approach the problem using vector components.
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
 
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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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