Angular Momentum of a rotating ball

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SUMMARY

The angular momentum of a 0.210-kg ball rotating at the end of a thin string with a radius of 1.35m and an angular speed of 10.4 rad/s is calculated using the formula L = Iω. The correct moment of inertia for a point mass is I = mr², which results in L = (0.210 kg)(1.35 m)²(10.4 rad/s). This calculation yields an angular momentum of 3.98 kg·m²/s, correcting the initial miscalculation of 1.592 kg·m²/s.

PREREQUISITES
  • Understanding of angular momentum and its formula L = Iω
  • Knowledge of moment of inertia for point masses, I = mr²
  • Familiarity with basic physics concepts such as mass, radius, and angular speed
  • Ability to perform calculations involving units of kg, m, and rad/s
NEXT STEPS
  • Study the derivation and applications of the moment of inertia for various shapes
  • Learn about the conservation of angular momentum in physical systems
  • Explore the differences between point masses and extended bodies in rotational dynamics
  • Investigate real-world applications of angular momentum in sports and engineering
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and anyone interested in understanding rotational dynamics and angular momentum calculations.

sammythesp3rmy
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Homework Statement


What is the angular momentum of a 0.210-kg ball rotating on the end of a thin string in a circle of radius 1.35m at an angular speed of 10.4 rad/s?


Homework Equations


I am using L = Iω


The Attempt at a Solution


I put I = (2/5)(0.210 kg)(1.35m)^2
ω = 10.4 rad / sec
When I try multiplying these out, I'm not getting the right answer. I come up with 1.592 kgm^2/sec and the answer is 3.98
 
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You're using the moment of inertia of a sphere, But it isn't the sphere that's spinning. The ball is essentially a point mass at the end of the string. The moment of inertia for a point mass is [itex]I=mr^2[/itex]. Try this and you will get the answer.
 

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