# How to find frequency by force method ?

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Derive the natural frequency $f_n$ of the system composed of two homogeneous circular cylinders , each of mass M, and the connecting link AB of mass m . Assume small oscillations . There is no slipping between cylinder and ground.  I got the frequency by the energy method , can we find it by force method ?
If yes , then how ? as there will be friction force and hinge force which are unknown .

asked May 6, 2018
Please show your attempt to use the force method! The question does not mention work done against friction, so you can ignore it. Your energy method solution did not include it.

The reaction forces at the hinge are more difficult. Draw FBDs for the cylinder and the link. The forces on the two cylinders are identical, so you need a diagram for only one. The hinge forces have horizontal and vertical components.  Apply Newton's 2nd law to each body.

As a further simplification, imagine that the cylinders overlap and become one. You can do this because their motion is identical. Then the link has zero length - ie it is only a point mass. (The question does not tell you the length of the link, only its mass. So the length is irrelevant, and could be zero.) Then you have one cylinder with mass 2M and an off-centre mass m, rocking on the table. Now there is no need to consider the forces at the hinge.
My try :  https://imgur.com/a/26s7Rjs

But in this how to find F$_2$ .

This answer uses a simplified method, which avoids the complications of having to deal with internal forces between the link and cylinders.

The motion of the two cylinders is identical, so they can be treated as one cylinder of mass $2M$. The link is then a point mass $m$ located at distance $r_0$ from the centre O. We can treat the link as a point mass because its orientation does not change throughout the motion. Like a point mass it has only translational KE, it has no rotational KE. The moment of inertia of the double-cylinder about the point of contact with the ground P is
$$I=2(\frac12Mr^2+Mr^2)=3Mr^2$$ The moment of inertia of the double-cylinder and link about P is therefore $$I_P=3Mr^2+mr_1^2 \approx 3Mr^2+m(r-r_0)^2$$ where $r_1$ is the distance LP and for small oscillations $r_1 \approx r-r_0$.

The torque about the axis at P is $\tau\approx mgr_0\theta$ in the small angle approximation. This acts to decrease $\theta$ so the equation of motion is $$\tau=-I_p \ddot \theta$$ $$mgr_0\theta \approx -[3Mr^2+m(r-r_0)^2]\ddot\theta$$ $$\ddot\theta+\frac{mgr_0}{3Mr^2+m(r-r_0)^2}\theta \approx0$$ The natural frequency is $$f_n\approx 2\pi \sqrt{\frac{mgr_0}{3Mr^2+m(r-r_0)^2}}$$

answered May 8, 2018 by (27,556 points)
selected May 8, 2018 by koolman
1 vote In your diagram the forces shown are those on the link, which is accelerating to the left while the cylinders accelerate to the right. For small oscillations the link has only horizontal acceleration; relative to point of contact P this is $(r-r_0)\alpha$ where $\alpha=\ddot\theta$ is the angular acceleration of the cylinders.

Applying Newton's 2nd Law to the link we have $$2F_2 \approx m(r-r_0)\alpha$$ $$2F_1\approx mg$$ The forces on the cylinders are equal and opposite to those on the link. For each cylinder the clockwise torque about P (the sense in which $\theta$ is decreasing) is $$\tau_P \approx r_0\theta F_1 + (r-r_0)F_2 \approx \frac12 m[gr_0\theta + (r-r_0)^2\alpha]$$
The equation of motion for each cylinder is $$-\tau_P=\frac32 Mr^2\alpha$$ $$-\frac12 m[gr_0\theta+(r-r_0)^2\alpha] \approx \frac32 Mr^2 \alpha$$ $$0 \approx mgr_0 \theta+ [3Mr^2 +m(r-r_0)^2]\alpha$$ $$\ddot \theta + \frac{mgr_0}{3Mr^2+m(r-r_0)^2}\theta \approx 0$$

answered May 7, 2018 by (27,556 points)
edited May 8, 2018
I think it should be $T=(r-r_0\cos\theta)F_2+(r_0\sin\theta)F_1=\frac32Mr^2\alpha$

But then also answer is not coming .
Yes, you are right. I have corrected my answer. Applying a consistent sign convention when there are internal forces can be quite confusing.