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Complicated Motion Problem

  • Thread starter PhDorBust
  • Start date
  • #1
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A garden roller with external radius a is pulled along a rough horizontal path by force F which acts at a point on its axle and is inclined at an angle alpha with the horizontal.

Given that resultant force is at a right angle to F, what is the acceleration of the roller? Express in terms of radius a, alpha, and radius of gyration k.

My set-up (and failure):

Let H be resultant force, f_s is frictional force

Forces in x-dir:
[tex]Hsin( \alpha) = Fcos( \alpha) - f_s[/tex]
Forces in y-dir:
[tex]-Hcos( \alpha) = Fsin( \alpha) - mg[/tex]
Rotational equation of motion:
[tex]af_s = \frac{mk^2}{a} \ddot{x}[/tex]
Assume no slipping:
[tex]\ddot{x} = a \ddot{ \theta}[/tex]
Solve for x-acceleration:
[tex]\ddot{x} = \frac{a^2 F - a^2 mgsin( \alpha)}{mk^2 cos( \alpha)}[/tex]
Correct answer:
[tex]\ddot{x} = \frac{ga^2 sin( \alpha) cos( \alpha)}{k^2 + a^2 sin^2 (\alpha)}[/tex]

Where am I going wrong? I don't see why F isn't in the answer...
 

Answers and Replies

  • #2
BruceW
Homework Helper
3,611
119
I haven't tried to work out the answer myself yet, but you're right that the 'correct answer' looks ridiculous.
It would predict the same acceleration even when F approaches zero, which is ridiculous.
 

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