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I completely understand the concepts behind uniform circular motion. But let's say you are spinning a ball connected by a string to a motor in a horizontal circle. When increasing the angular velocity of the spinning motor, I can't see how the ball connected to the string will have any force that allows it to increase its tangential velocity. How would the string be able to pull it so it accelerates tangentially all while undergoing circular motion? An example would be if you are swinging a ball in a circle above your head and you begin to spin it faster. How is the string causing the object to increase its linear speed? I believe tension can't cause it unless it's working at an angle less than 90 to the tangent because work must be done to increase the kinetic energy.

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    $\begingroup$ It depends. A stuntman might take the downward side of a looping track. A satellite in orbit might engage it's station-keeping ion drive. A wheel on a vehicle might get more power through the axle. In your string case the string will not be completely radial during acceleration (which implies minor variation from cicularity at the beginning and end of the accelerating phase). And so on. $\endgroup$ – dmckee Nov 21 '16 at 1:11
  • $\begingroup$ When you swing the ball over your head, and spin it faster, you can't do it if your hand doesn't move. Your hand has to move in a circle, pulling the ball tangentially (as @rob said) to higher speed. $\endgroup$ – Mike Dunlavey Nov 21 '16 at 1:38
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In the case that you describe, an individual swinging a mass horizontally on the end of a string, the string does not run directly to the centre of rotation. Instead, it runs to your hand, which in turn is moving in a circle about its centre of rotation . Sometimes the arm is involved, sometimes only a rotation at the wrist. ( Mime winding up a sling to throwing speed to see what I mean)

If everything is constant (and there's no drag on the mass), the line from mass to hand to centre of rotation is straight; the string tension exerts only the centripetal force needed to maintain the circle, as well as an upward component to keep the mass from dropping downward.

If you then speed up the circular motion of your hand to a new constant angular velocity, your hand's angular motion gets ahead of the mass's angular motion, and gets continuously farther ahead. So now the tension in the string is not in the line from the mass to the centre of rotation. There is a tangential component to the tension, constantly increasing, which serves to speed up the rotation of the mass.

This tangential force speeds up the mass's rotation, until the mass is rotating faster than your hand. Then the mass gets ahead, and your hand slows it down, and so on.

Picture a pendulum clock in a wheel-style space station, with the pendulum set to swing in the plane of the wheel, but currently at rest. Then speed up the space station slightly...

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  • $\begingroup$ I wondered (and I'm still not sure after your answer) whether you can accelerate an object with a string while it is on a perfectly circular trajectory. (I assume yes; the end point of the string must move in a manner which keeps the centripetal (radial) force at the proper value for the increasing speed. For example, at the start, when the object is at rest, the attachment point moves momentarily tangentially, i.e. exerts no radial force whatsoever.) $\endgroup$ – Peter - Reinstate Monica Nov 21 '16 at 10:19
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Uniform horizontal circular motion

Consider a bob of mass $m$ moving in uniform horizontal circular motion, at constant angular velocity $\omega=2\pi f$.

To maintain this motion a centripetal force $F_c$ has to be exerted on the bob:

$$F_c=mr\omega^2,\tag{1}$$

where $v=\omega r$, with $v$ the tangential speed.

We can determine $F_c$ from the force diagram, because is not moving either towards or away from the centre of rotation (no radial motion). The bob also does't move in the vertical direction. By Newton's second law this means:

$$T\sin\theta=mg\tag{2}$$

And of course:

$$F_c=T\cos\theta\tag{3}$$

Combining $(2)$ and $(3)$ we get:

$$F_c=mg\cot\theta $$

With $(1)$:

$$\begin{align}\frac{mv^2}{r}&=mg\cot\theta \\ \implies v^2 &=rg\cot\theta\end{align}$$

As we increase tangential speed, $\theta$ goes down and $r$ goes up. Reworking $(2)$:

$$T=\frac{mg}{\sin\theta}$$

So as the tangential speed goes up, $\theta$ diminishes and $T$ increases. Without this increase in tension $T$, the increase in $v$ would not be sustainable.

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    $\begingroup$ How does the tension T in your diagram, no matter how large it is, exert a tangential force? $\endgroup$ – DJohnM Nov 21 '16 at 3:50
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There must be tangential acceleration, from a component of the force parallel to the velocity.

Note that if your center-pointing force is different from the magic value $F_\text{centripetal} = mv^2/r$ for uniform circular motion, the radius of your circular motion will change, and the tangential velocity will change as well to conserve angular momentum.

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