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[I used calculus to solve this.]

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

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- Thread starter mrknowknow
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- #1

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[I used calculus to solve this.]

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

- #2

462chevelle

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subscribed

- #3

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

- #4

462chevelle

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

- #5

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

- #6

462chevelle

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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?

- #8

462chevelle

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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

- #9

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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?

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

- #10

462chevelle

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d=vt

That would be the only way i could think of relating distance, time, and velocity.

That would be the only way i could think of relating distance, time, and velocity.

- #11

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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?d=vt

That would be the only way i could think of relating distance, time, and velocity.

- #12

462chevelle

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- #13

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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?

- #14

462chevelle

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- #15

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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.

- #16

462chevelle

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

- #17

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Take it one step at a time, and post your answer to each, as far as you can go:would it be cosθ=vt i cant think of where to put my ordered pairs to be honest.

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?

- #18

462chevelle

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thanks

- #19

462chevelle

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

- #20

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How does gravity come into it?!θ=Vi(t)+(1/2)g(t^2) am i getting there with this equation?

Please post answers to each of the steps I laid out, as requested. Avoid wild guesses.

- #21

462chevelle

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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

- #22

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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.

- #23

462chevelle

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- #24

462chevelle

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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?

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