## samigupta8 Group Title find the magnitude of average accleration of the tip of second hand of length 10 cm during 10 seconds.... 10 months ago 10 months ago

1. Abhishek619 Group Title

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 Group Title

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

3. samigupta8 Group Title

@AllTehMaffs pls..hlp me in this question

4. Mashy Group Title

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 Group Title

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

6. Mashy Group Title

then what parth says is right

7. samigupta8 Group Title

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

8. samigupta8 Group Title

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

9. Mashy Group Title

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

10. samigupta8 Group Title

it is v^2/r

11. Mashy Group Title

there you ahve it.. plug it in!

12. samigupta8 Group Title

it's cuming as pi^2/90

13. samigupta8 Group Title

no no sory i made a mistake lemme solve it

14. samigupta8 Group Title

it's pi^2/9000

15. Mashy Group Title

yea.. thats right..

16. Mashy Group Title

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 Group Title

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

18. samigupta8 Group Title

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

19. samigupta8 Group Title

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

20. Vincent-Lyon.Fr Group Title

21. samigupta8 Group Title

ryt my answer is also that only

22. Vincent-Lyon.Fr Group Title

That's also the answer Mashy gave you.

23. Abhishek619 Group Title

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

24. samigupta8 Group Title

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

25. Vincent-Lyon.Fr Group Title

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 Group Title

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