From Angular Velocity to Angular Acceleration - How?

[berkeman]

From Angular Velocity to Angular Acceleration -- How?

Homework Statement

Calculate the vector acceleration as a function of angle for a mass rotating in a circle at the end of a string. The mass is rotating in the vertical plane, in the counterclockwise direction.

$$\hat{i}$$ and $$\hat{j}$$ are the unit vectors in the x and y directions, respectively.

Homework Equations

$$F = m \frac{v^2}{R} = m {\omega}^2 R$$

The Attempt at a Solution

This is my first thread in Intro Physics, so please be gentle

I'm helping a friend with this problem. We need to solve it in the general case, but for now, I'm just solving it for the minimum rotation velocity case. So at the top of the circle, the centriptal acceleration of the mass matches the acceleration due to gravity.

I've been able to derive the equation for the velocity vector as a function of the angle $$\psi$$ down from the vertical, but I'm not sure how to go from the velocity vector to the acceleration vector (both being functions of angle, not of time).

The equation for the velocity as a function of angle down from the vertical that I derived using TE = PE + KE for the mass is:

$$\vec{v}(\psi) = \hat{i} [\sqrt{2gR(1-cos\psi)} + \sqrt{gR}](-cos\psi) + \hat{j} [\sqrt{2gR(1-cos\psi)} + \sqrt{gR}](-sin\psi)$$

This gives the correct answers for the velocity at the top and bottom of the circle:

Top: $$\vec{v}(0) = -\hat{i} \sqrt{gR}$$

Bottom: $$\vec{v}(\pi) = \hat{i} 3\sqrt{gR}$$

But I'm having trouble figuring out how to go from the velocity vector $$\vec{v}(\psi)$$ to an acceleration vector $$\vec{a}(\psi)$$

It would seem that I need to convert the $$\vec{v}(\psi)$$ into a $$\vec{v}(t)$$, then differentiate, and convert $$\vec{a}(t)$$ into $$\vec{a}(\psi)$$, but that seems like a lot of work. Is there a trick or approach that I can use to shortcut that procedure?

Thanks for any hints or tutorial help!

 

[Delphi51]

Looks too complicated for me! I did play with it - considering only the magnitude of the velocity and beginning with your E = PE + KE, which for me worked out to
v^2 = gr(1 - 2cos(A)) where I am writing A instead of the Greek letter for convenience.

Differentiating that with respect to time gives
2v*dv/dt = 2gR*sin(A)*D=dA/dt
Using v = R*dA/dt this simplifies to
dv/dt = g*sin(A), which seems to make sense. It could be expressed in vector form as
a = g*sin(A)*cos(A) in the x direction and g*sin(A)*sin(A) in the y direction.

I'm probably vastly oversimplifying things - not knowledgeable about circular motion with non-constant speed.

 

[nlightner]

I would approach this problem by using the fundamental equations of circular motion to derive the relationship between angular velocity and angular acceleration. In this case, we can use the equation F = m \frac{v^2}{R} = m {\omega}^2 R, where F is the net force, m is the mass, v is the linear velocity, R is the radius of the circle, and \omega is the angular velocity.

To find the acceleration vector, we can use the definition of angular acceleration, \alpha = \frac{d\omega}{dt}, where \alpha is the angular acceleration and t is time. We can then use the chain rule to express this in terms of the angular velocity and the angle \psi, giving us \alpha = \frac{d\omega}{dt} = \frac{d\omega}{d\psi} \frac{d\psi}{dt} = \omega \frac{d\omega}{d\psi}.

Substituting this into the equation F = m {\omega}^2 R, we get F = mR \omega \frac{d\omega}{d\psi}. We can then use vector notation to express this as \vec{F} = mR \vec{\omega} \frac{d\vec{\omega}}{d\psi}, where \vec{F} is the net force, m is the mass, R is the radius, \vec{\omega} is the angular velocity vector, and \frac{d\vec{\omega}}{d\psi} is the derivative of the angular velocity vector with respect to the angle \psi.

From here, we can use the fact that the net force is equal to the sum of the centripetal force and the gravitational force (assuming the mass is rotating in a vertical plane), giving us \vec{F} = \vec{F}_{centripetal} + \vec{F}_{gravity}. We can then substitute in the expressions for these forces (mR \vec{\omega} \frac{d\vec{\omega}}{d\psi} = m \frac{v^2}{R} \hat{r} + m \vec{g}), where \hat{r} is the unit vector in the radial direction and \vec{g} is the gravitational acceleration vector.

After simplifying, we get the expression \vec{a} =