What is the Linear Speed of an Air Puck in Circular Motion?

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SUMMARY

The linear speed of an air puck in circular motion can be calculated using the relationship between tension, gravitational force, and centripetal acceleration. In this scenario, a 0.029 kg air puck revolves in a circle of radius 1.6 m, while a 1.4 kg mass provides the necessary tension. The correct formula to derive the linear speed (v) is derived from equating the forces: (1.4 kg * 9.81 m/s²) = (0.029 kg * v²) / (1.6 m). After resolving the algebra, the calculated linear speed is approximately 31.28 m/s, although initial calculations may yield errors due to algebraic mistakes.

PREREQUISITES
  • Understanding of centripetal acceleration and its formula: a = v²/r
  • Knowledge of free body diagrams and force equilibrium
  • Basic algebra skills for solving equations
  • Familiarity with gravitational force calculations: F = mg
NEXT STEPS
  • Review the concept of centripetal force and its applications in circular motion
  • Practice drawing and analyzing free body diagrams for various physical scenarios
  • Learn about the relationship between tension and gravitational force in systems involving pulleys
  • Explore common mistakes in algebraic manipulation and how to avoid them
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for practical examples of force analysis in rotational systems.

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Homework Statement


An air puck of mass 0.029 kg is tied to a string and allowed to revolve in a circle of radius 1.6m on a frictionless horizontal surface. The other end of the string passes through a hole
in the center of the surface, and a mass of 1.4kg is tied to it, as shown in the figure. The
suspended mass remains in equilibrium while the puck revolves on the surface. The acceleration of gravity is 9.81 m/s2 .
What is the linear speed of the puck? answer in m/s.
I got what the forces were- i drew a freebody diagram and had tension and gravity pointing down, with centrifigal acc pointing down as well.
I got tension to be 13.734N and gravity to be .28449N.
I plugged them into Fnet= ma
so T+Fg= m(centrif acc)
thing is, I'm not sure what to use for mass..is it the .029, the combined mass of the two or the mass of the weight on the bottom?
 
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Centripetal acceleration=v^2/r
T=(1.4 kg block)*g
T=(mass of small block)(centripetal acceleration) or T=(.029kg)v^2/r

Therefore, (1.4kg)*g=(.029Kg)v^2/r

I think you can take it from there.

With respect to deciding which masses go where: Think about which object*acceleration is supplying the force. The lower object is not spinning and is therefore not contributing to the "spinning force"
 
Also, I just noticed: If the object is falling near the surface of Earth the acceleration due to gravity is 9.81m/s^2. Or maybe I didn't understand part of your question.
 
Actually the more I look at this,

There are two free body diagrams to draw. The first one is of the block rotating around the table. The second is of the block underneath the table.

From that you will get two equations. Put them together as I did and the rest is algebra.
 
hmm i set them equal to each other and i got velocity to be 31.27982714m/s. However, when I submit that, it says I'm wrong...
 
That's not what I got for velocity.
(1.4kg)(9.81m/s^2)=(.029Kg)(v^2)/(1.6m)
 
ah faulty algebra on my part... Thanks for your help :)
 

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