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Assume a point object of mass $m$ slides along a hoop of radius $R$, starting from a position which makes 90 degrees with the line of radius connecting the center and the ground. Let the coefficient of kinetic friction between the hoop and the object be $\mu$. Assuming that the object starts at rest, what is the total work done by the friction when the object comes to the ground level?

My idea: the normal force at any instance is given by $$N=mg\sin\theta+\frac{mv^2}{R},$$ where $\theta$ is the angle between the radial line connecting the present position and the intital position of the object to the center of hoop. With this we have the frictional force as $$f_k=\mu\left(mg\sin\theta+\frac{mv^2}{R}\right),$$ so that the total work done by friction is $$W_k=\int_0^{\pi/2}\mu\left(mg\sin\theta+\frac{mv^2}{R}\right)R\mathop{\mathrm{d\theta}}.$$

The problem I am having is to figure out $v$ as a function of $\theta$, i.e $v(\theta)$. Any ideas?

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Mass sliding on ring.

Set up an equation of motion for the rotation of the mass around the centre-point.

$$\tau=I\alpha$$

Where:

$\tau=mg\cos\theta-\mu mg\sin\theta$

$I=mR^2$

$\alpha=\frac{d\omega}{dt}=\omega\frac{d\omega}{d\theta}$

So:

$$mg\cos\theta-\mu mg\sin\theta=mR^2\omega\frac{d\omega}{d\theta}$$

$$R^2\omega d\omega=g(\cos\theta-\mu \sin\theta)d\theta$$

Integrate between $0,\pi/2$ and $0,\omega$ to get an expression of $\omega^2$ in $\theta$. Then use $v=\omega R$.

Then calculate the gain in kinetic energy: $\Delta K=\frac{mv^2}{2}$ (*) and the loss in potential energy $\Delta U=mgR$. The difference between the two is the friction work.

(*) Or use $\Delta K=\frac{I\omega^2}{2}$.

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You are almost there. All you need to do now is realize that the velocity can be deduced from the kinetic energy, which is the difference between the potential energy lost and the work done by friction. And you have expressions for both of those. See if that gets you there.

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