Thursday 19 March 2020

homework and exercises - A rope wrapping around a cylinder




A massless rope of length $l$ is attached tangentially to a cylinder of radius $R$ and the free end of the rope is attached to a point mass $M$. When the rope is taut, the point mass is given an initial velocity $v_0$ perpendicular to the rope. All motion occurs on a horizontal frictionless surface.



How can we find the amount of time it takes for the point mass to reach the surface of the cylinder?



Answer



Okay, here's a simple solution then.


The force due to the tension of the rope is always perpendicular to the velocity, so it does not change the energy of the particle, and so the velocity is always $v_0$.


The angular velocity about the point of contact is $\omega=v_0/(l-R\theta)$ since $l-R\theta$ is the length between point of contact and particle as you pointed out.


The final angular position of the particle is $l/R$, so the total time $t$, $$t=\int^{l/R}_0\frac{dt}{d\theta}d\theta=\int^{l/R}_0\omega^{-1}d\theta=\frac{1}{v_0}\int^{l/R}_0(l-R\theta)d\theta=\frac{1}{v_0}(l(l/R)-\frac{R}{2}(l/R)^2)=\frac{l^2}{2Rv_0}$$


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