Non-Uniform Circular Motion - Find Change in time

[mcovalt]

Homework Statement

Find the time it takes for a particle initially at rest to travel around a circle with acceleration \ddot{\theta} = -3 \cos{\theta} to travel 1/4 of the circle.2. The attempt at a solution
\int_{0}^{{\frac{\pi}{2}}}-3 \cos{\theta}

Am I doing this right?

 

[ehild]

The integral of acceleration is velocity and the integral of velocity is displacement. You also need the initial angular velocity.

ehild

 

[mcovalt]

So I should do \int_{V_i}^{V_f}-3 \cos{\theta}

 

[ehild]

No, you have to integrate with respect to time, and this problem is not simple at all. Are you sure that you copied it correctly? The initial position is also needed.

ehild

 

[yster]

I don't think the initial position is required because you are given the initial velocity and the total angular displacement. I might be wrong but it sounds simple to me.

I do agree however that you have to integrate with respect to time because that is the link between acceleration, velocity and displacement.

You actually have to integrate twice

 

[mcovalt]

Integrating twice makes sense. I could graph the formula twice integrated and find the change in x for a quarter oscillation, correct?

Thanks for the help guys.

 

[mcovalt]

Oh boy. Does this require Elliptical integration?! If so, kill me now haha

 

[ehild]

I asked already, if you copied the problem correctly.

The usual method to solve such problems that contain only the unknown function and its second derivative with respect to time is to consider the unknown function as independent variable.

The angular velocity is ω=dθ/dt , so d²θ/dt²=dω/dt=(dω/dθ)ω ,

The original equation transforms to

0.5 d(ω²)/dθ =-3cos(θ),

which is easy to integrate, and you can get ω as function of the angle θ, which you should solve somehow but it is ugly any way.

ehild

 

[mcovalt]

Interesting. Thank you very much for the help. I did copy the question down correctly.