Tension and Centripetal Force in Circular Motion

[Lori]

Homework Statement

Where does T2cos(theta) come from ? Isn't mv^2/R the centripetal force which is the tension of rope 2?

Homework Equations

Fc = mv^2/R

3. Solution

Wait! The horizontal component of the circle is the centripetal force? So that part is mv^2/R?

I got confused and thought the tension of rope 2 is actually the centripetal force. But if the x point of that tension is the centripetal force, i see how they got T2Cos

since cos(theta) = Xcomp/T2
i get cos(theta) = (mv^2/R)/T2

so that T2cos(theta)= (mv^2/R)[/B]

 

[TSny]

Yes, the ball is moving in a horizontal circle as shown below. The centripetal force must point from the ball toward the center of the circle. You can see that the x-component of T₂ is the only force acting on the ball in this direction. So, the centripetal force is T₂cosθ.

 

[Lori]

Yes, the ball is moving in a horizontal circle as shown below. The centripetal force must point from the ball toward the center of the circle. You can see that the x-component of T₂ is the only force acting on the ball in this direction. So, the centripetal force is T₂cosθ.
View attachment 214370

Thank you. Your drawing really made sense of this all!