Question

find the magnitude of average accleration of the tip of second hand of length 10 cm during 10 seconds....

Answer 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.
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.

Answer 2

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

Answer 3

@AllTehMaffs pls..hlp me in this question

Answer 4

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 :-/

Answer 5

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

Answer 6

then what parth says is right

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

it is v^2/r

Answer 11

there you ahve it.. plug it in!

Answer 12

it's cuming as pi^2/90

Answer 13

no no sory i made a mistake lemme solve it

Answer 14

it's pi^2/9000

Answer 15

yea.. thats right..

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

The answer is π²/9000

Answer 21

ryt my answer is also that only

Answer 22

That's also the answer Mashy gave you.

Answer 23

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

Answer 24

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

Answer 25

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.



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