Solving Omega in Angular Velocity Equations: A Test Prep for Sarah

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SUMMARY

The discussion focuses on the application of omega (ω) in angular velocity equations, particularly in the context of preparing for a physics test. Key equations include θ = θ₀ + ωt + αt, where α represents angular acceleration. Participants emphasize the importance of substituting linear variables with their angular counterparts, such as replacing v with ω and a with α. The discussion also highlights the significance of rotational equations derived from standard kinematics equations.

PREREQUISITES
  • Understanding of basic kinematics equations
  • Familiarity with angular velocity concepts
  • Knowledge of angular acceleration (α)
  • Ability to manipulate equations involving rotational motion
NEXT STEPS
  • Study the derivation of rotational equations from linear kinematics
  • Learn about the relationship between linear and angular motion
  • Explore the implications of angular acceleration in real-world scenarios
  • Practice solving problems involving ω, α, and θ in various contexts
USEFUL FOR

Students preparing for physics exams, educators teaching rotational motion concepts, and anyone seeking to deepen their understanding of angular velocity equations.

sarahowen
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we are just about to take a test in physis.. I am quite scared. i have revised on all the topics required but i don't get the use of omega in angular velocity equations. what is the equation and when do i need to use it?
sarah
 
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well - u can take ur normal kinematics equation and replace v with omega.
like...

<br /> \theta = \theta_0 + \omega t + \alpha t<br />

alpha is the angular acceleration.
 
Just learn

v=rw

w= 2.pi.f = 2.pi / T

f = 1/T =w / 2.pi

a = v^2/r = rw^2

F=mv^2/r = mrw^2
 
Replace v by \omega

a by \alpha

and x by \theta

in your kinematics equations to get the rotational equations!

\omega = \omega_0 + \alpha t
\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2
\omega^2 = \omega_0^2 + 2\alpha\theta

And the Equations posted by Briton are also very important
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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