Here's the question you clicked on:
samigupta8
find the magnitude of average accleration of the tip of second hand of length 10 cm during 10 seconds....
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.
Radial acceleration is \(R\omega^2\) assuming constant angular velocity. The angular velocity is \(\dfrac{2\pi}{60} \rm rad s^{-1}\)
@AllTehMaffs pls..hlp me in this question
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 :-/
oh wait.. are you talking about centripetal acceleration!?
then what parth says is right
if it's right then how we r going to calculate acceleration
i m convinced with this that linear velocity is cuming as pi/300 m/sec
then u know what is the relation between linear vel and centripetal acceleration
there you ahve it.. plug it in!
it's cuming as pi^2/90
no no sory i made a mistake lemme solve it
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
bt the answer is given as pi/3000 m/s^2
here have they taken the value of one pi as 3 and then solve it
if dey hve done like this then it wud give the same answer to our question also am i ryt....?
The answer is π²/9000
ryt my answer is also that only
That's also the answer Mashy gave you.
the answer given is correct. it's pi/3000 m/s^2. at least this is what i'm getting.
how r u getting this answer cuz i m getting it as pi^2/9000
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
drats.. thanks a lot for that update man!!