centripital acceleration

[Whiteshoes]

Lets say Im spinning a yoyo above my head in a horizontal circle. if the centripetal acceleration is the net force acting on the yoyo and will point directly towards the center (towards my hand holding the string), then what about the force of gravity downwards? what is canceling out the downward force of gravity if the tension in the string is only pointing in the X direction and gravity is in the Y direction.

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[Whiteshoes]

ok i know that the tension also has a y component TSin(theta) which cancels out the mg..but what if Im spinning it so fast that yoyo does not dip down with respect to my hand and theta is 0 degrees. Tsin(theta) = 0 then whats canceling out the mg in the downward direction?

thanks

 

[ljc]

ok i know that the tension also has a y component TSin(theta) which cancels out the mg..but what if Im spinning it so fast that yoyo does not dip down with respect to my hand and theta is 0 degrees. Tsin(theta) = 0 then whats canceling out the mg in the downward direction?

thanks

Won't happen. You are actually spinning it with some vertical component, regardless of how it feels. Mg will not go away. This is why if you try to spin the yo-yo at a greater than parallel (to the ground) level, it will fall downward. Mg and the vertical component of centripetal force are working in the same direction.
In order for it to remain in flight, you must be pulling upwards ever so slightly.

 

[milski]

Won't happen. You are actually spinning it with some vertical component, regardless of how it feels. Mg will not go away. This is why if you try to spin the yo-yo at a greater than parallel (to the ground) level, it will fall downward. Mg and the vertical component of centripetal force are working in the same direction.
In order for it to remain in flight, you must be pulling upwards ever so slightly.

There is no vertical component, you can do that quite easily by tying the rope to some sort of carousel which will not give you any vertical movement. The vertical component of the tension is T*sin(theta). As you spin the object faster, you increase T and decrease theta. While you can make theta very small, it will never become zero. In other words, it will not spin in a horizontal circle. In reality, you can make theta small enough so that you won't notice the slightly conical shape of the circle.

 

[ljc]

There is no vertical component,

As you spin the object faster, you increase T and decrease theta. While you can make theta very small, it will never become zero.

I think we're misunderstanding each other. Because theta will not reach zero, there is always SOME vertical component, necessary to overcome mg.

 

[milski]

I think we're misunderstanding each other. Because theta will not reach zero, there is always SOME vertical component, necessary to overcome mg.

Yes, I did misunderstood you, sorry. I thought you meant that you're giving some sort of vertical bumps to object when rotating it by hand. We do agree that it will be slightly below a horizontal plane.

 

[Whiteshoes]

thanks!