Solving Part A & B of a Circular Motion Homework Problem

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SUMMARY

The discussion focuses on solving a circular motion problem involving a 0.5 kg ball attached to an 80 cm cord, which is whirled at a constant speed. For Part A, the calculated tensions at the top and bottom of the circle are 92.7 N and 102.7 N, respectively, which differ from the expected values of 93.8 N and 104 N. In Part B, the participant incorrectly concludes that the time for one complete revolution is 0.4 seconds, while the correct answer is 1.8 seconds. The formulas used include T=m(g+v²/r) for the bottom and T=m(g-v²/r) for the top of the circle.

PREREQUISITES
  • Understanding of circular motion dynamics
  • Familiarity with the equations of motion: v=2πR/T
  • Knowledge of tension calculations in circular motion
  • Basic physics concepts such as mass, gravity, and centripetal acceleration
NEXT STEPS
  • Review the derivation of the centripetal acceleration formula
  • Learn how to calculate tension in different scenarios of circular motion
  • Study the implications of varying tension on circular motion
  • Explore the relationship between period, frequency, and angular velocity in circular motion
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Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for examples of tension calculations in practical applications.

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Homework Statement

Part A : A 0.5 Kg ball is attached to a cord 80 cm long. The ball is whirled in a vertical circle at a constant speed. If the ball takes 0.4 seconds to go around once, what is the tension in the cord at the top and bottom of the circle?

Part B How long does it take for the ball to go around once if the tension in the cord at the top of the circle is zero?

Homework Equations



v=2Pi*R/T
T=m(g+v^2/r) (bottom of the circle)
T=m(g-v^2/r) (top of the circle)

The Attempt at a Solution



For part A i used both of the circular motion formulas to fin the tension and i got totally different answers from what it suppose to be for top and bottom. Top it suppose to be 93.8N and bottom it suppose to be 104 N i got 92.7N top and 102.7 N bottom. ( I do not know if the answer is a print error).

For part B I tried to use v=2Pi*R/T to get the time and i keep getting the same time in part A which is 0.4 secs. But the answer is 1.8 seconds.

Can someone explain to be what I'm doing wrong? thanks!
 
Last edited:
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ScienceGeek24 said:

Homework Statement




Part A : A 0.5 Kg ball is attached to a cord 80 cm long. The ball is whirled in a vertical circle at a constant speed. If the ball takes 0.4 seconds to go around once, what is the tension in the cord at the top and bottom of the circle?

Part B How long does it take for the ball to go around once if the tension in the cord at the top of the circle is zero?

Homework Equations



v=2Pi*R/T
T=m(g+v^2/r) (bottom of the circle)
T=m(g-v^2/r) (top of the circle)


The Attempt at a Solution



For part A i used both of the circular motion formulas to fin the tension and i got totally different answers from what it suppose to be for top and bottom. Top it suppose to be 93.8N and bottom it suppose to be 104 N i got 92.7N top and 102.7 N bottom. ( I do not know if the answer is a print error).

For part A I tried to use v=2Pi*R/T to get the time and i keep getting the same time in part A which is 0.4 secs. But the answer is 1.8 seconds.

Can someone explain to be what I'm doing wrong? thanks!


Your formulas are correct, can you post your exact calculations for it?
 
m=0.5 kg
r=0.8 m
T=0.4 seconds
v=12.5 m/s

Bottom part of the circle:

0.5kg(12.5^2m/s/0.8m+9.8m/s^2)= 102.7N

Top part of the circle

0.5kg(12.5^2m/s/0.8m-9.8m/s^2)=92.7N
 
Last edited:

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