# Theory on tangential acceleration [closed]

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.

• ask your ta ..? – agentp Apr 23 '17 at 2:03
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• Are you really thinking, that you copy-paste your homework here and the answering machine here will solve it for you? – peterh Apr 24 '17 at 5:07
• If it is not homework or an exam question, then why is it worth 6 marks? – Solar Mike Apr 24 '17 at 7:12

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?

• 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 ? – Amber Apr 23 '17 at 16:05
• 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. – Amber Apr 23 '17 at 18:29
• 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. – MrYouMath Apr 23 '17 at 19:14
• root((8/10)*2-1)=v Is this how I plug in the 8 ? – Amber Apr 23 '17 at 20:03
• no, I got 0.77. and thats not correct – Amber Apr 23 '17 at 20:22