b) I dont understand how to take the integration like its integral of ft from 0 to 2 pi = integral of velocity, and the integral of the first is the area under the graph= 8j and then I dont know how to get the answer fron there cause dont get what to plug in ? Any help is greatly appresiated.

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    $\begingroup$ ask your ta ..? $\endgroup$ – agentp Apr 23 '17 at 2:03
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    $\begingroup$ Welcome to Engineering! This looks like a homework question. In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please edit your question to include this information. $\endgroup$ – Wasabi Apr 23 '17 at 5:54
  • $\begingroup$ Are you really thinking, that you copy-paste your homework here and the answering machine here will solve it for you? $\endgroup$ – peterh Apr 24 '17 at 5:07
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For the first question, you can split up the equations of motion into normal and tangential direction:

$$ma_n=F_n$$ $$ma_t=F_t.$$

The tangential force $F_t=12/\pi \text{ N}$. Hence, $a_t=\frac{12}{10\pi} \frac{\text{m}}{\text{s}^2}$. For the normal force (centripetal force) you have to remember that $F_n=m\frac{v^2}{r}$. Hence, $a_n=\frac{v^2}{r}=1\frac{\text{m}}{\text{s}^2}$. The total acceleration is given by $a=\sqrt{a_t^2+a_n^2}=1.07\frac{\text{m}}{\text{s}^2}$.

So, now it's up to you. Try the other questions and tell us what you have tried or where you got stuck.

Edit: For b) we note that $$ma_t=F_t\implies \frac{F_t}{m}=a_t=\dfrac{dv}{dt}.$$ This can be rewritten by the chain rule (Note, that $ds/dt=v$) as: $$\frac{F_t}{m}=\dfrac{dv}{ds}\dfrac{ds}{dt}=v\dfrac{dv}{ds}\implies \int_{s=0}^{2\pi}F_t ds=m\int_{v_0}^{v_{2\pi}}vdv.$$

The integral of on the right-hand side is just the area in your $s-F_t$-diagram. It turns out to be $8 \text{ J}$. Can you continue from here?

  • $\begingroup$ what I did for part b was make an equation for the graph and replace a with dv*v/ds and integrated to get velocity but that was not correct. Am I not on the right track ? $\endgroup$ – Amber Apr 23 '17 at 16:05
  • $\begingroup$ The integral on the right with velocity is the area of s-Ft. How I thought we had to find velocity from the integration? and for Ft I made the equation 36/pi^2(s-5pi/3) and integrated is this correct. $\endgroup$ – Amber Apr 23 '17 at 18:29
  • $\begingroup$ I dont think your integral is right. You should get 8 J. Remember that $F_t$ is piecewise defined. It is very unpracticable to express it as one function. Simply calculate the area by positve rectangle, positive triangle, negative triangle and negative rectangle. $\endgroup$ – MrYouMath Apr 23 '17 at 19:14
  • $\begingroup$ root((8/10)*2-1)=v Is this how I plug in the 8 ? $\endgroup$ – Amber Apr 23 '17 at 20:03
  • $\begingroup$ no, I got 0.77. and thats not correct $\endgroup$ – Amber Apr 23 '17 at 20:22

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