What Is the New Friction Force if Angular Speed Is Doubled?

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SUMMARY

The discussion centers on calculating the new friction force when the angular speed of a spinning cylinder is doubled. The frictional force is initially defined as F_f = μ_k · F_N, where F_N is the normal force. The conclusion reached is that while F_f is proportional to ω², the actual friction force required to keep a person against the wall remains equal to their weight, which does not change with increased angular speed. Therefore, the correct answer is that the new friction force does not simply double or quadruple but remains constant.

PREREQUISITES
  • Understanding of angular speed and its relationship to centripetal force
  • Knowledge of frictional force and normal force concepts
  • Familiarity with the equations of motion in circular dynamics
  • Basic grasp of the coefficient of friction (μ_k)
NEXT STEPS
  • Study the relationship between angular speed and centripetal acceleration in detail
  • Learn about the maximum static friction force and its implications in rotational motion
  • Explore the concept of normal force in various physical scenarios
  • Investigate real-world applications of friction in rotating systems, such as amusement park rides
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Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as educators looking for examples of friction in circular motion.

vaio-911
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Homework Statement



A barrel of fun consists of a large vertical
cylinder that spins about the vertical axis.
When it spins fast enough, any person inside
will be held up against the wall.

http://i.snag.gy/Wh1B9.jpg

Question:

If you double the angular speed (i.e., \omega _2=2\cdot \omega _1) what is the new friction force f_2?

Answer choices: 2\cdot F_1,1/3,1/4,1/2,3,1.5,1/3.5,1,2.5,4

Homework Equations



F_f=\mu_k\cdot F_N

v=r\cdot \omega

a_c=\dfrac{v^2}{r}

The Attempt at a Solution



The frictional force is defined as F_f=\mu_k\cdot F_N.

The normal force of the object is the centripetal force that is pointing towards the center of the cylinder (perpendicular to surface).

Thus frictional force is

\begin{align*}F_f&amp;=\mu_k\cdot (m\cdot a_c)\\&amp;=\mu_k\cdot \left(m\cdot \dfrac{v^2}{r}\right)\\<br /> &amp;=\mu_k\cdot\left(m\cdot\dfrac{(r\cdot\omega)^2}{r}\right)<br /> \end{align*}

This means that \boxed{F_f\propto \omega^2}

so F_2=4\cdot F_1, but when I submitted it, it said that I was wrong.

What am I doing wrong?
 
Last edited:
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Hi vaio-911. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

vaio-911 said:
The frictional force is defined as F_f=\mu_k\cdot F_N.
I don't think it is, actually. I think that is the maximum frictional force before relative motion can occur. Isn't that so?

The actual frictional force to secure each person against gravity is their weight, and that doesn't change just because you spin them faster!

Maybe this is a trick questionhttp://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon5.gif Is your professor a barrel of fun? :smile:
 
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