a stone attached on a rope. That stone performs a uniform circular motion with a radius of 1 ~ m The stone makes one revolution per second. The average acceleration of the stone during 1/4 revolution is?

Question

anonymous · Answer

@Tushara

anonymous · Answer

is the answer 39.5ms^-2 towards the centres of the circle?

anonymous · Answer

$$\Pi \sqrt{128}m/s ^{2}$$

anonymous · Answer

ill let you know if i figure it out

anonymous · Answer

okay!(:

anonymous · Answer

are sure there was no time mentioned in the question because if it wasnt then i m getting same answer as tushara using v^2/r

experimentx · Answer

what kind of circle is it??
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anonymous · Answer

there is no diagram, just those words..

experimentx · Answer

are you sure the answer is $ \pi \sqrt{128} $ ??
it is said that it's " uniform circular motion "
the acceleration is always constant in uniform circular motion

anonymous · Answer

yes, i am sure.. i translated from dutch to english so yes, everything is the way its meant to be unless you want me to write the dutch version. :)

experimentx · Answer

I'm pretty sure ... they all above got the right answer. unless if you got the 
left figure (vertical circle - which is not likely (large angle pendulum))
or you are missing something like ... the motion begins with rest and reaches to speed of 1rev/sec

anonymous · Answer

and I am pretty sure that that's the question and answer..

anonymous · Answer

@experimentX ....this is right question..

anonymous · Answer

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