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

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Abhishek619
 one year ago
Best ResponseYou've already chosen the best response.1there 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Radial acceleration is \(R\omega^2\) assuming constant angular velocity. The angular velocity is \(\dfrac{2\pi}{60} \rm rad s^{1}\)

samigupta8
 one year ago
Best ResponseYou've already chosen the best response.0@AllTehMaffs pls..hlp me in this question

Mashy
 one year ago
Best ResponseYou've already chosen the best response.2ahhww.. 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 :/

Mashy
 one year ago
Best ResponseYou've already chosen the best response.2oh wait.. are you talking about centripetal acceleration!?

Mashy
 one year ago
Best ResponseYou've already chosen the best response.2then what parth says is right

samigupta8
 one year ago
Best ResponseYou've already chosen the best response.0if it's right then how we r going to calculate acceleration

samigupta8
 one year ago
Best ResponseYou've already chosen the best response.0i m convinced with this that linear velocity is cuming as pi/300 m/sec

Mashy
 one year ago
Best ResponseYou've already chosen the best response.2then u know what is the relation between linear vel and centripetal acceleration

Mashy
 one year ago
Best ResponseYou've already chosen the best response.2there you ahve it.. plug it in!

samigupta8
 one year ago
Best ResponseYou've already chosen the best response.0it's cuming as pi^2/90

samigupta8
 one year ago
Best ResponseYou've already chosen the best response.0no no sory i made a mistake lemme solve it

Mashy
 one year ago
Best ResponseYou've already chosen the best response.2u 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

samigupta8
 one year ago
Best ResponseYou've already chosen the best response.0bt the answer is given as pi/3000 m/s^2

samigupta8
 one year ago
Best ResponseYou've already chosen the best response.0here have they taken the value of one pi as 3 and then solve it

samigupta8
 one year ago
Best ResponseYou've already chosen the best response.0if dey hve done like this then it wud give the same answer to our question also am i ryt....?

VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.1The answer is π²/9000

samigupta8
 one year ago
Best ResponseYou've already chosen the best response.0ryt my answer is also that only

VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.1That's also the answer Mashy gave you.

Abhishek619
 one year ago
Best ResponseYou've already chosen the best response.1the answer given is correct. it's pi/3000 m/s^2. at least this is what i'm getting.

samigupta8
 one year ago
Best ResponseYou've already chosen the best response.0how r u getting this answer cuz i m getting it as pi^2/9000

VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.1Hi @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: \(\vec a_{average} = \Large \frac {\vec v_2\vec v_1}{t_2t_1}\) with \(\Delta t = t_2t_1\)=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

Mashy
 one year ago
Best ResponseYou've already chosen the best response.2drats.. thanks a lot for that update man!!
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