Vertical Circle question

  • #1
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Homework Statement


The questions asks for the centripetal acceleration of the mass in the vertical circle shown in the picture. It also states the object has exactly enough velocity to maintain a vertical circle.

Homework Equations


F=ma, a=v^2/r

The Attempt at a Solution


I've tried solving for the velocity by setting mg=m*v^2/r and then subtracting mgcos(114) to get the velocity and then plugging it back into a=v^2/r, but to no avail.
 

Attachments

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Answers and Replies

  • #2
gneill
Mentor
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Hi hittce, Welcome to Physics Forums.

The velocity that you solved for is the critical velocity at the top of the circle to just maintain the circular motion (it's the critical point where the object is moving at its slowest speed). At any other location on the circle the speed will be greater as the object trades gravitational PE for KE...
 
  • #3
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I actually tried solving it that way before I posted and got it wrong, but upon reading your response I decided to try again and it turns out I made a small error before. Anyway, now I got the correct answer, so thanks for taking the time to reply.
 
  • #4
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The next part of the question asks for the total acceleration. I got the centripetal acceleration to be 21.43, and then I found the tangential acceleration to be 3.986 by taking g*sin(24) and then I proceeded to take the square root of the sum of the squares to get 21.796 since the two accelerations were perpendicular, but the answer was wrong. Any tips as to where I went wrong?
 
  • #5
Nathanael
Homework Helper
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I found the tangential acceleration to be 3.986 by taking g*sin(24)
I usually check the special cases of 0° and 90° when using sine/cosines. So in this case, if the angle was 90° instead of 114°, then you're saying that the tangential acceleration would be g*sin(0)=0?
 
  • #6
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Ah, so I should have left it as sin(114) after all, so much for my drawings. But was I right in thinking that once I have this number that I find the total acceleration by taking the square root of the sum of the squares of tangential and centripetal acceleration?
 
  • #7
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Never mind, I was impatient and went ahead and submitted the answer and it was correct. Thanks for the help guys.
 
  • #8
Nathanael
Homework Helper
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I was thinking cos(24°) but that's the same thing as sin(114°). Any way you like best works. And yes you can use pythagorean's theorem to add them because, like you said, they are perpendicular.
 

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