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A sphere of neglectable radius is placed on a very long and frictionless rod (which we can approximate to a straight line) on which it is able to move. The rod rotates around one of its end points with constant angular velocity $\omega$. Find:

  1. Whether the sphere gets closer or farther from the fixed end point
  2. The direction and orientation of the total force acting on the particle in an inertial frame of reference (e.g. that whose origin lies on the fixed end point of the rod)
  3. The equation of motion of the particle in the same reference frame.

I don't know how to answer 1. I might say it moves towards the centre, but I have poor physical intuition.

As for 2, the only force acting on the particle is the vincular reaction of the rod, which constrains the particle to move over the rod itself. But there is no explicit formula for vincular forces: their value can be found only if the total acceleration is known, which is after all what we're trying to evaluate in the first place...

Perhaps conservation of energy or angular momentum could help? Constraint forces do no work, so the potential energy would be null at all times...but of course we know nothing about kinetic energy, and the same for momentum. The only thing we know is that $\theta (t)=\omega t$, where $\theta$ is the angle between the $x$ axis and the position vector. Thus, we have one of two polar coordinates. How to find the other one, however, I have no clue.

Could you give me a hint on how to approach this problem?

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2 Answers 2

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  1. Imagine that you are on a car, and that it turns sharp. What force will you experience ? Will you go toward the center of the virage, or the exterior ? This situation is the same as the sphere on the rod.
  2. Here, you just have to give the direction and orientation of the force. There is no friction, however friction is defined as the tangential component of the reaction of the rod, so the reaction of the rod must be normal to the rod. Plus, real forces are not altered by changing of reference frame, so the reaction of the rod will always be perpendicular to the rod. Now, considering the motion of the sphere, you can find the direction and the orientation of the force.
  3. It will be easier to study the sphere in the non-inertial frame of reference of the rod. To do this, you have to write the second law of Newton in an inertial frame of reference, and transform acceleration in order to make the acceleration of the sphere in the rod frame appear. You should get something looking like $m\mathbf{a_{rod\, frame}} = \mathbf{F} + \mathbf{something}$
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  • $\begingroup$ Why is the force perpendicular to the rod? $\endgroup$
    – Adrian
    May 14, 2017 at 21:24
  • $\begingroup$ The reaction $\mathbf{R}$ of the rod can be written as $\mathbf{R} = \mathbf{R}_N+\mathbf{R}_T$ where $\mathbf{R}_N$ (resp. $\mathbf{R}_T$) is its normal (resp. tangential) component. However, there is no friction, ie. $\mathbf{R}_T = 0$ $\endgroup$
    – Spirine
    May 14, 2017 at 21:57
  • $\begingroup$ I don't quite understand. Why does friction only have a tangential component? Couldn't the reaction impressed by the rod have a both components? I know that the constraint force (minus friction) has to be normal to the trajectory, but the segment (the rod) is not the trajectory itself, at least not in the intertial frame. $\endgroup$
    – Adrian
    May 15, 2017 at 3:29
  • $\begingroup$ The rod reaction does, indeed, have both a radial and azimuthal component. The azimuthal component is called the normal reaction, the radial component is the friction, assumed here to be 0. $\endgroup$
    – john
    Dec 24, 2023 at 14:33
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I don't have the full solution, but you asked for a hint.

Start with the equation for acceleration in polar coordinates

$$a=(\ddot r - r \dot \theta ^2) e_r+(r \ddot \theta+2 \dot r \dot \theta) e_\theta$$

For constant angular velocity this becomes

$$a=(\ddot r - r \dot \theta ^2) e_r+2 \dot r \dot \theta e_\theta$$

In the absence of friction there is no force along the rod, just a normal reaction, N $$N e_\theta=m(\ddot r - r \dot \theta ^2) e_r+2m \dot r \dot \theta e_\theta$$

The r component gives

$$ \ddot r = r \dot \theta ^2$$

The $\theta$ component gives

$$N =2m \dot r \dot \theta$$

Remember that $\dot \theta$ is a known constant

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  • $\begingroup$ The radial equation is stuctured like the SHM equation with the "restoring" force directed away from the center. $\endgroup$
    – john
    Dec 24, 2023 at 14:43

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