Register to reply

Kinematic questions(velocity of 2 contact points)

by hihiip201
Tags: contact, kinematic, points, questionsvelocity
Share this thread:
hihiip201
#1
Feb28-13, 11:58 PM
P: 163
in the picture below



http://tinypic.com/view.php?pic=9riq9l&s=6


if we know the angular velocity of the gear on the left, as well as their current distance from contact point, what can we say about the velocity of the right rod at the contact point at anytime t?


we know it is not just the velocity of the contact point of the left rod because the right rod can be sliding along the left rod. but how do we find its exact velocity?
Phys.Org News Partner Physics news on Phys.org
First in-situ images of void collapse in explosives
The first supercomputer simulations of 'spin?orbit' forces between neutrons and protons in an atomic nucleus
Magnets for fusion energy: A revolutionary manufacturing method developed
Andrew Mason
#2
Mar1-13, 05:42 AM
Sci Advisor
HW Helper
P: 6,654
Quote Quote by hihiip201 View Post
in the picture below



http://tinypic.com/view.php?pic=9riq9l&s=6


if we know the angular velocity of the gear on the left, as well as their current distance from contact point, what can we say about the velocity of the right rod at the contact point at anytime t?


we know it is not just the velocity of the contact point of the left rod because the right rod can be sliding along the left rod. but how do we find its exact velocity?
If you know the angular velocity of the left rod and the distance of the contact point from the centre of the left gear (rl), you can determine the velocity of the contact point (ie. in terms of ωl and rl). You then just have to relate that to the velocity of the contact point on the right rod. That relationship changes at the point where the two rods are parallel (contact point is the whole rod). So you would have to break it down into two parts (ie. one where the end of the right rod contacts the left and the other where the end of the left rod contacts the right).

AM
hihiip201
#3
Mar1-13, 01:50 PM
P: 163
Quote Quote by Andrew Mason View Post
If you know the angular velocity of the left rod and the distance of the contact point from the centre of the left gear (rl), you can determine the velocity of the contact point (ie. in terms of ωl and rl). You then just have to relate that to the velocity of the contact point on the right rod. That relationship changes at the point where the two rods are parallel (contact point is the whole rod). So you would have to break it down into two parts (ie. one where the end of the right rod contacts the left and the other where the end of the left rod contacts the right).

AM

how exactly do we relate the velocity of the contact point of left rod and right rod?

we know the velocity of left rod at contact point, but certainly right rod's velocity at the contact point wouldn't be the same since it is sliding along the left rod right?

Andrew Mason
#4
Mar2-13, 09:48 PM
Sci Advisor
HW Helper
P: 6,654
Kinematic questions(velocity of 2 contact points)

Quote Quote by hihiip201 View Post
how exactly do we relate the velocity of the contact point of left rod and right rod?

we know the velocity of left rod at contact point, but certainly right rod's velocity at the contact point wouldn't be the same since it is sliding along the left rod right?
You have to work it out from the geometry.

Consider the situation shown with the tip of the right rod is contacting the body of the left rod. You have to relate the velocity of the tip of the right rod to the velocity of the point that it is in contact with:

ωrLr = ωlrl + v where v is the speed of the end of the right rod along the length of the left rod (ie. drl/dt)/

I haven't worked it out but I think that is how you have to do it. The trick is to express v in terms of the angular velocities of the two rods.

AM
hihiip201
#5
Mar3-13, 12:38 AM
P: 163
Quote Quote by Andrew Mason View Post
You have to work it out from the geometry.

Consider the situation shown with the tip of the right rod is contacting the body of the left rod. You have to relate the velocity of the tip of the right rod to the velocity of the point that it is in contact with:

ωrLr = ωlrl + v where v is the speed of the end of the right rod along the length of the left rod (ie. drl/dt)/

I haven't worked it out but I think that is how you have to do it. The trick is to express v in terms of the angular velocities of the two rods.

AM


my first guess is also the equation you have above, but what i have trouble is that, should vector v in your equation be based on the geometry of the two rods at current time t? or t = t +dt? to be it makes more sense to use the time at t+dt or else the left rod will end up in a awkward position.
Andrew Mason
#6
Mar3-13, 09:14 AM
Sci Advisor
HW Helper
P: 6,654
Quote Quote by hihiip201 View Post
my first guess is also the equation you have above, but what i have trouble is that, should vector v in your equation be based on the geometry of the two rods at current time t? or t = t +dt? to be it makes more sense to use the time at t+dt or else the left rod will end up in a awkward position.
v is just the component of the tangential velocity of the end of the right rod that is in the direction of the radius vector rl. So it is based on the angle between the two rods.

AM
hihiip201
#7
Mar3-13, 11:32 AM
P: 163
Quote Quote by Andrew Mason View Post
v is just the component of the tangential velocity of the end of the right rod that is in the direction of the radius vector rl. So it is based on the angle between the two rods.

AM

right but the angle between two rods at the current time? or some dt later??


if I break the velocity into two component, the component that is along the contact point of left rod , and the velocity tangential to the left rod.

After some dt later, the position of the contact point of my right rod would be a the position of the contact point of the left rod, minus some distance away from the center of the left rod.


I don't know if my question make sense, sorry about the confusion.
Andrew Mason
#8
Mar3-13, 01:23 PM
Sci Advisor
HW Helper
P: 6,654
Quote Quote by hihiip201 View Post
right but the angle between two rods at the current time? or some dt later??
It does not matter. dt is arbitrarily small. If you set v(t) at t then at t+dt it will be v(t+dt) = v + dv. If you set θ(t) at t then at t+dt it will be θ(t+dt) = θ+dθ = θ+ω(dt)

if I break the velocity into two component, the component that is along the contact point of left rod , and the velocity tangential to the left rod.

After some dt later, the position of the contact point of my right rod would be a the position of the contact point of the left rod, minus some distance away from the center of the left rod.
What you want to find is the velocity of the contact point on the left rod. Can you express that in terms of the velocity of the end of the right rod? ie. break down the tangential velocity of the right rod ωrrr into components that are in the direction of the rl and perpendicular to that direction (ie tangential to rl).

AM


Register to reply

Related Discussions
Applying force to a flat object with contact points Introductory Physics Homework 4
2D kinematic questions Introductory Physics Homework 5
Near and far points with contact lenses Introductory Physics Homework 2
Weight distribution, multiple points of contact Classical Physics 1
Distance between polygons in 3d? (points of contact as a function of time) General Math 4