A disc of radius r is spinning about its center in the horizontal plane with a constant angular speed w (omega). A bug walks along the radius of the spinning disc traveling from the center of the disc toward the edge. The bug maintains a constant speed v relative to the disc. (In other words, if the disc were not spinning, the bug would travel at a speed v relative to the disc and ground.) What is the vector acceleration and (if not already given) the magnitude of the acceleration of the bug relative to the ground in terms of the three variables: r, omega, v. Show all work in the derivation.

[I used calculus to solve this.]

If someone could lead me in the right direction I'd appreciate it.

462chevelle
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haruspex
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Can you write down the location of the bug at time t in XY coordinates?

462chevelle
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since he hasn't responded can you help me through this problem?

haruspex
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since he hasn't responded can you help me through this problem?
Sure. Can you answer my post #3?

462chevelle
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not with certainty, at t=0, I would say the bug is at (0,0). I cant think of how I could find it at t=1. since it has a constant velocity it would seem that the vector acceleration is equal to the v/t. this is challenging for me having only variables in the problem.

haruspex
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not with certainty, at t=0, I would say the bug is at (0,0). I cant think of how I could find it at t=1. since it has a constant velocity it would seem that the vector acceleration is equal to the v/t. this is challenging for me having only variables in the problem.
For the sake of argument (it won't affect the answers to the OP), suppose the bug starts heading along the positive x axis (at speed v). Where would it be at time t if the disc were stationary?
Suppose it would then have reached point A of the disc. Start again with the disc rotating now. After time t, will the bug be at point A of the disc?

462chevelle
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time would be distance/velocity
are we looking to put this into function notation?
like Δd(v,Δt)=vΔt
if the disc is not spinning there no derivitave calculation needed right? or do i need to find the derivation of distance in terms of time? ill take this one step at a time. and see if im on the right track before i attempt to solve with it spinning.
thanks

haruspex
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time would be distance/velocity
are we looking to put this into function notation?
like Δd(v,Δt)=vΔt
if the disc is not spinning there no derivitave calculation needed right? or do i need to find the derivation of distance in terms of time? ill take this one step at a time. and see if im on the right track before i attempt to solve with it spinning.
thanks
You don't need any calculus for what I've asked so far. The bug moves at constant speed v, so how far has it gone in time t?

462chevelle
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d=vt
That would be the only way i could think of relating distance, time, and velocity.

haruspex
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d=vt
That would be the only way i could think of relating distance, time, and velocity.
Right. Now suppose that would have brought the bug to point A on a staionary disc. Will the bug still reach point A after time to if the disc is rotating? If so, where will point A be in the ground XY coordinates?

462chevelle
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if it is rotating and the velocity of the bug is constant i would expect the bug to be at point A's radial distance from the center. it should be at (x-0)^2+(y-0)^2=A^2 all of the points at distance A relative to the ground.

haruspex
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if it is rotating and the velocity of the bug is constant i would expect the bug to be at point A's radial distance from the center. it should be at (x-0)^2+(y-0)^2=A^2 all of the points at distance A relative to the ground.
It will be at one of those points. The disc is rotating at constant rate w, so if A started at (vt, 0), where will it be at time t?

462chevelle
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can i get a hint. i would think i would need to get the derivative of something so i know what the rate of change on the curve is but i cant think of how to write it.

haruspex
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can i get a hint. i would think i would need to get the derivative of something so i know what the rate of change on the curve is but i cant think of how to write it.
Consider a rod OA of length 1, initially with end O at (0,0) and end A at (1,0). It is rotated about the origin at constant rate w for time t. What angle has it rotated through? Where is end A now? No calculus needed - this is just trig.

462chevelle
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would it be cosθ=vt i cant think of where to put my ordered pairs to be honest.

haruspex
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would it be cosθ=vt i cant think of where to put my ordered pairs to be honest.
Take it one step at a time, and post your answer to each, as far as you can go:
If the rod OA of length 1 rotates at rate w for time t, what angle does it rotate through?
If the rod OA of length 1 rotates (anticlockwise) through angle theta, what is the x coordinate of A?
If the rod OA of length 1 rotates (anticlockwise) through angle theta, what is the y coordinate of A?
If the rod OA of length 1 rotates at rate w for time t, what are the x and y coordinates of A?
If a rod OA of length r rotates at rate w for time t, what are the x and y coordinates of A?
If a rod of initially zero length lengthens at constant rate v, how long is it at time t?
If a rod OA of initially zero length (and notionally at angle 0 to the +ve x axis) lengthens at constant rate v while also rotating anticlockwise at rate w, what are the x and y coordinates of A at time t?

462chevelle
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im not ignoring this, just not much time with school at the moment. but it will probably take me till Saturday to do some research to try and answer these questions. I will let you know when I get some time

thanks

462chevelle
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θ=Vi(t)+(1/2)g(t^2) am i getting there with this equation?

haruspex
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θ=Vi(t)+(1/2)g(t^2) am i getting there with this equation?
How does gravity come into it?!
Please post answers to each of the steps I laid out, as requested. Avoid wild guesses.

462chevelle
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i didn't mean to put gravity there. sorry but the equation for the angle of rotation I have in my notes is
θ=Vi(t)+(1/2)∂(t)^2
∂= angular acceleration
but I don't know where length of the rod would come into play with that.
or would it be θf=θi+Vi(t)+(1/2)∂(t)^2

haruspex
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i didn't mean to put gravity there. sorry but the equation for the angle of rotation I have in my notes is
θ=Vi(t)+(1/2)∂(t)^2
∂= angular acceleration
but I don't know where length of the rod would come into play with that.
or would it be θf=θi+Vi(t)+(1/2)∂(t)^2

That's a very unusual and confusing choice of symbols. Maybe you copied it down wrongly. I would expect to see ##\theta_f = \theta_i + \omega_i t + \frac{1}{2}\alpha t^2##.
In the present case, rotation rate is constant.

462chevelle
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i just haven't figured out latex yet. those are just the closest symbols I could see there. so there is no accell. ill have to try again.

462chevelle
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ive looked and looked at this I guess im at a brick wall. the only way I could think of doing it is to make an isosceles triangle where a=1, b=1, and c=(2)^1/2
then that still don't tell me anything. this question is a bit over my head but I cant quit thinking about it.
it has to be either a concept im not in full understanding going from math to physics or something I don't even know yet. since this isn't my homework problem can I get another hint and I will see if I can get anywhere with it?