Circular Motion Homework: Tension & Min. Speed

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SUMMARY

The discussion focuses on calculating the tension in a cord during uniform-speed circular motion of a 0.50 kg ball with a radius of 2.0 m and a speed of 10 m/s. The tension at the lowest point is determined to be 30 N using the equation T = m(Ac + g), where Ac is the radial acceleration. Additionally, the minimum speed for the ball to maintain circular motion is calculated to be 4.4 m/s, derived from the condition where tension becomes zero at the top of the circular path. The analysis emphasizes the importance of considering the angle of the string with the vertical to accurately determine the point of zero tension.

PREREQUISITES
  • Understanding of uniform circular motion principles
  • Knowledge of radial acceleration calculations
  • Familiarity with Newton's second law (Fnet = ma)
  • Basic trigonometry related to angles in circular motion
NEXT STEPS
  • Study the concept of radial acceleration in circular motion
  • Learn how to derive tension equations in vertical circular motion
  • Explore projectile motion principles following circular motion
  • Investigate the effects of varying mass and radius on tension and speed
USEFUL FOR

Physics students, educators, and anyone studying dynamics of circular motion, particularly in the context of forces and tension in strings.

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Homework Statement



A small ball of mass 0.50 kg is attached to a cord and perform uniform-speed circular motion of radius 2.0 m in a vertical plane.

i) If the speed of the circular motion is 10m/s, determine the tension in the cord at the lowest point of the circular motion.
ii) Determine the minimum possible speed of this circular motion.

Homework Equations



radical acceleration = v^2/r
Fnet=ma

The Attempt at a Solution



i) T-mg=mAc, where Ac is the radical acceleration.
T=mAc+mg=m(Ac+g)=0.5*(10^2/2+9.81)=30N

ii) At the top position,
mg-T=mAc
T=0 for minimum speed,
mg=mAc=m(v^2/r)
v=√(gr)=√(9.81*2)=4.4m/s
 
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Your attempt at the first bit of the question is correct. In the second bit of the question you considered the top most point of the vertical circle to be the point where tension is zero but as the mass is attached to a flexible chord the tension becomes zero before the ball reaches to the top and hence the ball leaves the circle and goes into a projectile motion. Try framing your equation as a function of the angle made by the string with the vertical then locate the point where tension becomes zero.
 

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