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## samigupta8 2 years ago find the magnitude of average accleration of the tip of second hand of length 10 cm during 10 seconds....

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1. Abhishek619

there are some assumptions which you should make is that, the seconds have move with constant angular velocity 'w' time period is 2pi/w=60. w=2pi/60 10 secs means it travels 1/6th part of the clock.|dw:1382102831264:dw| it goes from A to B. change in velocity and solve like the previous case how we did it. and then that divide by the total time taken=10secs. do this and you'll get the ans.

2. ParthKohli

Radial acceleration is $$R\omega^2$$ assuming constant angular velocity. The angular velocity is $$\dfrac{2\pi}{60} \rm rad s^{-1}$$

3. samigupta8

@AllTehMaffs pls..hlp me in this question

4. Mashy

ahhww.. i dunno how to even begin.. average acceleration shoudl be zero cause before 10 seconds and after 10 seconds, the second hand comes to stop :-/

5. Mashy

oh wait.. are you talking about centripetal acceleration!?

6. Mashy

then what parth says is right

7. samigupta8

if it's right then how we r going to calculate acceleration

8. samigupta8

i m convinced with this that linear velocity is cuming as pi/300 m/sec

9. Mashy

then u know what is the relation between linear vel and centripetal acceleration

10. samigupta8

it is v^2/r

11. Mashy

there you ahve it.. plug it in!

12. samigupta8

it's cuming as pi^2/90

13. samigupta8

no no sory i made a mistake lemme solve it

14. samigupta8

it's pi^2/9000

15. Mashy

yea.. thats right..

16. Mashy

u knw u can solve this simply.. 60 seconds --- 2 pi rads 1 second -- pi/30 rads so a = w^2 r = (pi^2/900)*0.1

17. samigupta8

bt the answer is given as pi/3000 m/s^2

18. samigupta8

here have they taken the value of one pi as 3 and then solve it

19. samigupta8

if dey hve done like this then it wud give the same answer to our question also am i ryt....?

20. Vincent-Lyon.Fr

The answer is π²/9000

21. samigupta8

ryt my answer is also that only

22. Vincent-Lyon.Fr

That's also the answer Mashy gave you.

23. Abhishek619

the answer given is correct. it's pi/3000 m/s^2. at least this is what i'm getting.

24. samigupta8

how r u getting this answer cuz i m getting it as pi^2/9000

25. Vincent-Lyon.Fr

Hi @samigupta8, I am so sorry, but @Mashy and myself have led you to a wrong solution to your problem. Actually, @Abhishek619 gave you the right answer at the very beginning. As the problem asks for AVERAGE acceleration from position 1 to position 2, as acceleration is a vector, the right answer is: $$\vec a_{average} = \Large \frac {\vec v_2-\vec v_1}{t_2-t_1}$$ with $$\Delta t = t_2-t_1$$=10 s and $$\vec v_1$$ and $$\vec v_2$$ as drawn in Abhishek619's sketch. |dw:1382428975483:dw| Hence $$\vec v_2-\vec v_1$$ has the same magnitude v as v1 and v2 because all angles are π/3 the magnitude of the average acceleration is a = v/$$\Delta t$$ Since v = 2πr/60s = 2π x 0.1 /60 = π / 300 m/s and $$\Delta t$$ = 10 s then average acceleration is a = π / 3000 m/s²

26. Mashy

drats.. thanks a lot for that update man!!

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