Need help explaining this Centripetal force problem

[Lori]

Homework Statement

Homework Equations

T = mv^2/r
Fc = mv^2/r

The Attempt at a Solution

I'm aware that at the bottom of this motion, that the tension force is greater than the weight. And that the tension force at the top is weaker so we added T + mg = mv^2/R. for th bottom, it's T-mg = mv^2/R

Why is the centripetal force on the sides, in this example, just T + mgcos(60)? Can someone explain how this is the case

 

[rude man]

You should write one equation describing this motion. Keeping in mind that in order for the mass to be circling as shown the net force on the mass must always be m v^2/r, write the equation accordingly. Then you will have T(θ) for any θ. But you need a second equation to get v. To obtain v you need to use the additional info given to you: T(30°) and v(30°).

 

[I like Serena]

Hey Lori,

The force of gravity is not in the same direction as the required centripetal force.
So we break up the force of gravity into 2 components, one that is centripetal, and one that is parallel with the motion.
The force of tension and the component of gravity that is centripetal must add up to the required centripetal force $\frac{mv^2}{r}$.
The component of gravity that is parallel with the motion is responsible for increasing the speed.

 

[Lori]

Hey Lori,

The force of gravity is not in the same direction as the required centripetal force.
So we break up the force of gravity into 2 components, one that is centripetal, and one that is parallel with the motion.
The force of tension and the component of gravity that is centripetal must add up to the required centripetal force $\frac{mv^2}{r}$.
The component of gravity that is parallel with the motion is responsible for increasing the speed.

thanks, this make sense! I guess i never had to break force of gravity into components for the top and the bottom motion on a circle cause force of gravity is already vertical with centripetal force. Never had a problem like this until now.

 

[Lori]

Hey Lori,

The force of gravity is not in the same direction as the required centripetal force.
So we break up the force of gravity into 2 components, one that is centripetal, and one that is parallel with the motion.
The force of tension and the component of gravity that is centripetal must add up to the required centripetal force $\frac{mv^2}{r}$.
The component of gravity that is parallel with the motion is responsible for increasing the speed.

I have one question though. I found the same answer using mgsin30 +20 = mv^2/R since Fg is in the direction of centripetal force and T force is given in direction of Fc already. But why is the work shown in this problem mgcos(60) ? Isn't this the direction of gravity vertically?

 

[I like Serena]

I have one question though. I found the same answer using mgsin30 +20 = mv^2/R since Fg is in the direction of centripetal force and T force is given in direction of Fc already. But why is the work shown in this problem mgcos(60) ? Isn't this the direction of gravity vertically?

Don't we have that sin 30^o=cos 60^o?

And yes, Fg is always vertically.
The angle of Fg with the string is 60^o, therefore we get Fg cos 60^o as the centripetal component.
Alternatively, the angle of Fg with the velocity is 30^o, therefore the centripetal component is also Fg sin 30^o, which is the same.
It all depends on which triangle we pick to calculate the centripetal component.