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I was trying to use Newton's second law to describe the motion of the following pendulum:

enter image description here

However, I was confused as to how to include the spring in Newton's second law. I was able to set up a differential equation when I used torque, but I was unable to solve it if I use only F = ma.

My attempt:

enter image description here

I was not sure how to incorporate the spring into the equation.

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  • $\begingroup$ This problem is not easily solvable equating forces because there is an unknown force at the point of suspension of the pendulum that prevents it from moving axially. Even if you include that after drawing free-body analysis the problem will just become much more complicated. Using torques about the suspension is the better approach. $\endgroup$ – biryani Feb 16 '16 at 5:51
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Is $\theta$ very small ? Then you can just use $F=kx$ in the horizontal direction. It is also unclear what is the rest length of the spring. Let's assume when the bob is in the mean position, the spring is in its rest length.

Then when swinging leftwards (as in fig in OP), the spring will stretch by $x=L\sin\theta$; and hence a horizontal force $F=kL\sin\theta$ will act towards right due to spring, helping the $T\sin\theta$.

If $\theta$ is not small, then you should use rotational mechanics, centred on the point of suspension, as suggested by biryani. You could also break the forces into radial and tangential components to make it easier.

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