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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....
 9 months ago
 9 months ago
samigupta8 Group Title
find the magnitude of average accleration of the tip of second hand of length 10 cm during 10 seconds....
 9 months ago
 9 months 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.
 9 months 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 \(\dfrac{2\pi}{60} \rm rad s^{1}\)
 9 months ago

samigupta8 Group TitleBest ResponseYou've already chosen the best response.0
@AllTehMaffs pls..hlp me in this question
 9 months 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 :/
 9 months ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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