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daniel_i_l

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I'm trying to wrap my head around angular momentum using the following setup:

There's a ball on a frictionless table that's connected to a string. The other end of the string is connected to a hole in the table. In the initial position the ball spins around the hole.

Now, we start pulling the string through the hole and thus shortening the amount that's connected to the ball.

From conservation of angular momentum the ball is now spinning faster because it's radius is smaller. Now, angular momentum is conserved because the force of the string only pulls the ball towards the center. But in that case, what causes the angular velocity to increase?

I think that the answer lies in the fact that even though the force is radial, it still causes acceleration in the direction of the spin via the Coriolis acceleration. Assuming that the speed with which the ball moves towards the center is constant [tex]v_{c}[/tex] then the angular acceleration caused by the Coriolis acceleration is [tex] a(r) = -2 \omega v_{c} = -\frac{2 v(r) v_{c}}{r} [/tex]

This is a differential equation of [tex] v(r) [/tex] whose solution is [tex] v(r) = \frac{v_{0} r_{0}}{r} => v(r)r = v_{0}r_{0}[/tex] where [tex] v_{0} [/tex] and [tex] r_{0} [/tex] are the initial velocity in the direction of the spinning and the radius. This equation is exactly the conservation of angular momentum.

So apparently, the conservation of angular momentum is just another way of saying that forces at right angles to the velocity still change the velocity because of Coriolis acceleration.

Is my conclusion correct? If so, how can I solve the differential equation when [tex]v_{c}[/tex] changes with [tex]r[/tex]?

Thanks

There's a ball on a frictionless table that's connected to a string. The other end of the string is connected to a hole in the table. In the initial position the ball spins around the hole.

Now, we start pulling the string through the hole and thus shortening the amount that's connected to the ball.

From conservation of angular momentum the ball is now spinning faster because it's radius is smaller. Now, angular momentum is conserved because the force of the string only pulls the ball towards the center. But in that case, what causes the angular velocity to increase?

I think that the answer lies in the fact that even though the force is radial, it still causes acceleration in the direction of the spin via the Coriolis acceleration. Assuming that the speed with which the ball moves towards the center is constant [tex]v_{c}[/tex] then the angular acceleration caused by the Coriolis acceleration is [tex] a(r) = -2 \omega v_{c} = -\frac{2 v(r) v_{c}}{r} [/tex]

This is a differential equation of [tex] v(r) [/tex] whose solution is [tex] v(r) = \frac{v_{0} r_{0}}{r} => v(r)r = v_{0}r_{0}[/tex] where [tex] v_{0} [/tex] and [tex] r_{0} [/tex] are the initial velocity in the direction of the spinning and the radius. This equation is exactly the conservation of angular momentum.

So apparently, the conservation of angular momentum is just another way of saying that forces at right angles to the velocity still change the velocity because of Coriolis acceleration.

Is my conclusion correct? If so, how can I solve the differential equation when [tex]v_{c}[/tex] changes with [tex]r[/tex]?

Thanks

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