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

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Abhishek619 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

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

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

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

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

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

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

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

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

samigupta8 Group TitleBest ResponseYou've already chosen the best response.0
it is v^2/r
 one year ago

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

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

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

samigupta8 Group TitleBest ResponseYou've already chosen the best response.0
it's pi^2/9000
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.2
yea.. thats right..
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

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

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

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

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

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

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

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

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

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.1
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: \(\vec a_{average} = \Large \frac {\vec v_2\vec v_1}{t_2t_1}\) 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
 one year ago

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