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

1. anonymous

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. anonymous

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. anonymous

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

6. anonymous

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. anonymous

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

10. samigupta8

it is v^2/r

11. anonymous

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. anonymous

yea.. thats right..

16. anonymous

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. anonymous

21. samigupta8

ryt my answer is also that only

22. anonymous

That's also the answer Mashy gave you.

23. anonymous

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. anonymous

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. anonymous

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