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How much time goes by before the merry-go-round is turning 0.5 revolutions per second

  • Thread starter Carnivroar
  • Start date
  • #1
128
1
Torque = 2m * 50N
M.o.I = (400kg * 2m^2)/2
a = 1/8 rad/s^2 = 0.019894368 revs/s^2

I don't even know which formula to use

But I tried this one

ω= ω0 + αt

0.5rev/s = 0 + 0.019894368rev/s^2 * t

t = 23.132

That's seems like a lot of time... not sure if it's right.
 

Answers and Replies

  • #2
92
0


You're nearly there. ω is the angular velocity, you have the revolutions per second. Now, if I remember correctly,

ω=2∏η
where ω is angular velocity
η is revolutions per second
 
  • #3
128
1


Ooops, I mean 25.132, that was a typo. Is that the correct answer?
 
  • #4
92
0


Yes, that is the answer I get, just got there in a different way.
 
  • #5
128
1


Yes, that is the answer I get, just got there in a different way.
Good then

I figured it out, my mistake was thinking that the 50N push was just one push and then release, so I was confused as to why the merry-go-round would accelerate.

But it's a constant 50 N force, which is why it's accelerating, did not realize that at first.

Thanks
 

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