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Magnitude/direction of the ball's acceleration

  • Thread starter 14h54
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  • #1
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Homework Statement



A 2.0 kg ball swings in a vertical circle on the end of an 80 cm-long string. The tension in the string is 20 N when its angle from the highest point on the circle is theta = 30.

A) What is the ball's speed when theta = 30?
B) What is the magnitude of the ball's acceleration when theta = 30?
C) What is the direction of the ball's acceleration when theta = 30? Give the direction as an angle from the r-axis.

Homework Equations


f=ma; a = v2/r

The Attempt at a Solution



A) T-mgcos(theta)=mv2/r; v=sqrt((t-gcos(theta))*r) which is 3.8 m/s (the correct answer)

I have no idea how to go about finding part b and c. The only thing I may know about part c is that I'll have to use arctan at some point.

Thanks for all the help, I really appreciate it :)
 

Answers and Replies

  • #2
114
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It may be easier to solve for part c first, and then part b. Finding the acceleration vector may be the best thing you can do for yourself right now. I recommend drawing a picture of the force diagram when the ball is at 30 degrees. Will the ball's acceleration always be radially directed outward, always directed along the circular path of the pendulum's motion, or always down with the gravitational force?
 
  • #3
114
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And then after finding the direction, square each component, add them, and then square root the sum for magnitude! Hope this helps!
 

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