Cylinder pulled by rope on a rough surface

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SUMMARY

The discussion centers on calculating the acceleration of a cylinder being pulled by a force F at an angle θ on a rough surface. The relevant equations include the moment of inertia \(I = kMR^2\) and the relationship \(M = I\alpha\). The derived formula for acceleration is \(a = \frac{F(R\cos\alpha - r)}{MR(1+k)}\), which accounts for the torque and the moment of inertia based on the rotation axis. The solution emphasizes the importance of specifying the rotation axis when applying the moment of inertia.

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Karol
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Homework Statement


A rope is pulled off with a force F from a combined cylinder. the smaller cylinder's radius is r and the big one's is R. the cylinder rotates on a rough surface without sliding. the force makes an angle θ with the horizon.
What's the acceleration's "a" magnitude and direction.

Homework Equations


Rigid body: ##M=I\alpha##
$$I=kMR^2$$

The Attempt at a Solution


Torque relative to the contact point, the torque's arm is:
$$\left( R-\frac{r}{\cos\alpha} \right)\cos\alpha=R\cos\alpha-r$$
$$M=I\alpha\rightarrow F(R\cos\alpha-r)=kMR^2\cdot \alpha,\; R\alpha=a$$
$$\Rightarrow a=R\frac{F(R\cos\alpha-r)}{kMR^2}=\frac{F(R\cos\alpha-r)}{kMR}$$
 

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The moment of inertia depends on the position of the rotation axis. You have to specify it. Usually I=kMR2 is used with respect to the centre of the rolling body.
 
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$$a=R\frac{F(R\cos\alpha-r)}{MR^2(1+k)}=\frac{F(R\cos\alpha-r)}{MR(1+k)}$$
 

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