Help Finding velocity of an object in circular motion at an angle.

[Adam17]

Homework Statement

A 0.160 kg ball attached to a light cord is swung in a vertical circle of radius 70.0 cm. At the top of the swing, the speed of the ball is 3.26 m/s. The centre of the circle is 1.50 m above the floor. What is the velocity if the ball is 30 degrees below the horizontal

Homework Equations

The Attempt at a Solution

Was going to use this equation,T=mv^2/r - mg Sin() but I don't have the tension, and I don't know how to find it at and angle. So there must be another way that I am not thinking of.

 

[cepheid]

How about conservation of energy?

 

[Adam17]

So find the Potential Energy by mghcos(theta) and the Kinetic energy will be equal and opposite to it? Then derive the Kenetic energy equation to find velocity?

 

[cepheid]

So find the Potential Energy by mghcos(theta) and the Kinetic energy will be equal and opposite to it? Then derive the Kenetic energy equation to find velocity?

NO. They will not be equal and opposite. It's just that total energy is conserved, so their sum will remain constant. You know what both the kinetic energy and the potential energy are at the top of the swing. Therefore you know the total energy. You also know what the potential energy is at any point on the circle. So you should be able to find the kinetic energy at any point, since their sum remains constant.

 

[Adam17]

Ohhh ok, thanks for the clarification on the conservation, though the title is pretty self explanatory lol . I was interpreting Ep=-Ek in the wrong sense. So, I can find the ep at any point in the circle by mg(hcos theta), find the Ek by subtracting Ep from Me and then Solve for Velocity sqrt(2Ek/m), right?

 

[cepheid]

Ohhh ok, thanks for the clarification on the conservation, though the title is pretty self explanatory lol . I was interpreting Ep=-Ek in the wrong sense. So, I can find the ep at any point in the circle by mg(hcos theta), find the Ek by subtracting Ep from Me and then Solve for Velocity sqrt(2Ek/m), right?

It sounds about right, assuming that what you call Me is the mechanical energy of the system (the same as the total energy, in this case): the sum of the kinetic and the potential energies.

As for the equation you mentioned, you were probably thinking of the equation:

ΔEp = -ΔEk

The deltas are important. The change in potential energy is equal to the negative of the change in kinetic energy. (So that the net change in energy is zero -- energy conservation). This is true in the case where gravity is the only force that does work. So what it's saying is that if you find the change in potential energy between any two points on the circle, it will be equal to the negative of the change in kinetic energy between those two points. This is equivalent to what I told you. To see this, say we evaluate the energy of the system at two points: point 1 and point 2. The equation above says that:

Ep2 - Ep1 = -(Ek2 - Ek1) = Ek1 - Ek2

Rearranging this, we get:

Ep2 + Ek2 = Ep1 + Ek1

Which is what I told you in my previous post. The sum remains the same at any point in the circle, which is an equivalent statement to the statement that the net change in energy is zero, since an change in kinetic energy is canceled out by a change in potential energy of opposite sign.

 

[Adam17]

Thanks for your help Cepheid, I am taking physics through correspondance and It becomes very painful when I miss something from the content and you have eased the pain!