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In my Physics course we are dealing with uniform circular motion. In this scenario there is an object in horizontal uniform circular motion. enter image description here

In the FBD provided in the problem the gravitational force acting on the ball is cancelled by a Normal Force.

enter image description here Where did this normal force come from?

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There is none; the free-body diagram is wrong. In actuality, what will happen is that the rope will make a small angle but non-zero angle with the horizontal, so that there is some vertical component of the tension that cancels out gravity. The remaining horizontal component of the force will provide the needed centripetal acceleration.

enter image description here

It is not too hard to show that the faster the object goes in the circle, the closer the angle will be to the horizontal. In the present case, with the given values of $v$, $R$, and $g$, it can be shown that the angle the rope makes with the horizontal is less than 5° (try working out the precise angle yourself!) This means that it's a pretty good approximation, as far as the tension is concerned, to treat the rope as though it's horizontal; and if you solved the problem this way, you'd end up an answer that's pretty close to the answer you'd get using the correct free-body diagram. But of course, that doesn't make the given free-body diagram right.

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  • $\begingroup$ Appreciate the the help man. One of the best thought-out explanations I have ever gotten on this site :) $\endgroup$
    – snow_razer
    Mar 23, 2021 at 18:23
  • $\begingroup$ Could you explain how I can find the angle the rope makes with the horizontal? $\endgroup$
    – snow_razer
    Mar 23, 2021 at 18:39
  • $\begingroup$ @snow_razer: It's probably better that you figure it out yourself. But roughly speaking, you do what you always do: write down Newton's Law for the sum of the forces in the horizontal direction, write down Newton's Law for the sum of the forces in the vertical direction, and solve the resulting system of equations for the unknowns you're interested in. $\endgroup$ Mar 23, 2021 at 18:46
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The diagram is at best misleading. If the object is being swung in mid-air the string cannot be horizontal. $F_N$ must be the vertical component of the tension in the string, and $F_T$ must be the horizontal component of the tension.

Alternatively, the diagram is completely wrong, and the object is not being swung in mid-air at all but is on a horizontal surface. In this case the string is indeed horizontal, $F_N$ is the normal force from the surface and $F_T$ is the tension in the string.

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