Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Rigid body/ force applied on a wheel

  1. Dec 11, 2008 #1


    User Avatar
    Gold Member

    Hi PF,
    I'm a bit disappointed because the final exam is coming up next Monday and I can't solve this problem. Not that I don't know how to proceed, but I just get a wrong answer.

    1. The problem statement, all variables and given/known data
    See figure to imagine the situation. This is about a wheel of radius R which contains a circle of radius r in its middle. A cord is enrolled over the circle and a force is exerted on the cord.
    We observe that there's a critical angle (over the horizontal, of the force) such that the wheel roll in a different sense whether the angle is lesser than the critical angle or greater than it. Find the critical angle.

    2. Relevant equations
    In reality what happens is that I must find to what's equal the static friction force and equaling it to 0N. Whether it's negative or positive the wheel roll in a sense or the other.

    3. The attempt at a solution
    First I chose a center of momentum as being the point of application of the force applied on the circle.
    I notice that the wheel suffer a rotation and a translation. So I can say that the equations of motion are different from 0. Precisely [tex]\frac{dP}{dt}=Ma_{cm}=F_E=Fcos \theta +F_{\text {friction}}[/tex].
    And [tex]\frac{dL}{dt}=M_{externals}=M_{\text {frictional force}}=\vec r _{F_{\text {friction}}} \wedge \vec F_{\text {friction}}=[-r\sin \theta i -(R-r\cos \theta)j] \wedge F_{\text{friction}}i=(R-rcos \theta )F_{\text{friction}}k[/tex].
    Why did I wrote that the momentums are only the momentum of the frictional force? Because according to my center of momentum, the momentum of the weight of the wheel cancels out with the normal reaction of the ground, and the momentum of the force applied on the circle is simply worth 0 because it is precisely applied on my center of momentums.

    As it is a rotation and a translation, [tex]L_{\text {my center of momentum}}=L_{\text {orbital}}+L_{\text {spin}}[/tex]. From which I can say [tex]\frac{dL}{dt}=[/tex][tex]\frac{dL_{\text {orbital}}}{dt}[/tex][tex]+\frac{dL_{\text {spin}}} {dt}[/tex].
    Now comes the big calculus... I'll not go through details.
    I got that [tex]\frac{dL_{\text {orbital}}}{dt}=-Mr \cos \theta a_{cm} k[/tex]. And [tex]\frac{dL_{\text {spin}}}{dt}=I\alpha =\frac{MR^2 \alpha }{2}[/tex]. I must find [tex]\alpha[/tex].
    With the condition of rolling, [tex]\alpha=-\frac{a_{cm}}{R}[/tex]. Thus [tex]\frac{dL_{\text {spin}}}{dt}=-\frac{MRa_{cm}}{2}k[/tex].
    Now I can write [tex]L_{\text {my center of momentum}}=(R-rcos \theta )F_{\text{friction}}k=-Mr\cos \theta a_c{cm}k-\frac{MRa_{cm}}{2}k \Leftrightarrow F_{\text {friction}}=-Ma_{cm}(r\cos \theta +\frac{R}{2})(\frac{1}{R-r\cos \theta})[/tex], replacing [tex]a_{cm}[/tex] by its value I finally get that [tex]F_{\text {friction}}=-M(\frac{F\cos \theta +F_{\text{friction}}}{})(r\cos \theta +\frac{R}{2})(\frac{1}{R-r\cos \theta}) \Leftrightarrow F_{\text{friction}}=- \left( Fr\cos ^2 \theta+\frac{FR\cos \theta}{2} \right ) \left ( \frac{2}{(2+2\cos \theta)+R)(R-r\cos \theta)} \right )[/tex], which I believe is always negative, meaning that the frictional force never changes of direction no matter what [tex]\theta[/tex] is worth. I made at least a mistake...

    Attached Files:

    Last edited: Dec 11, 2008
  2. jcsd
  3. Dec 11, 2008 #2


    User Avatar
    Gold Member

    Nevermind! I just realized that as the wheel is moving, the moment of the force applied on the circle is not worth 0 after having started to move.
    Well, I'll try again.
    EDIT: Hmm, but the way I did it should work. It doesn't matter that the system will evolve. I described it initially, and the frictional force should have its definitive direction instantly after the wheel started to roll.
    Last edited: Dec 11, 2008
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook