# Calculate a derivation of a the angular speed of a solid body

#### luimucar

I am practicing for a mechanics exam.

Please find the problem and solution in the image:

I am not able to reach the solution.

The question is to derivate an expression of the angular speed to obtain the angular acceleration.

The angular speed is a function of the rotation angle, so the problem, I think, it is a differential equation thing. So I tried with all derivation rules (implicit and so on), but I am unable to reach the solution.

How do I derivate that?

Thank you for the help

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#### benit13

Yes, you get a 1st order ODE.

Some hints...

We have

$$\displaystyle \dot{\phi} = \frac{2}{R} \sqrt{\frac{(M-\mu m_B gR)\phi}{(3m_A + 2m_B)}}$$

If we let

$$\displaystyle A = \frac{2}{R} \sqrt{\frac{(M-\mu m_B gR)}{(3m_A + 2m_B)}}$$

we can reduce the formula to

$$\displaystyle \dot{\phi} = A\phi^{1/2}$$

This makes the ensuing algebra a bit more manageable. Then you can then perform another substitution of the form

$$\displaystyle u = \phi^{1/2}$$

$$\displaystyle \frac{du}{d \phi} = \frac{d\left(\phi^{1/2}\right)}{d\phi} = \frac{1}{2}\phi^{-1/2} = \frac{1}{2u}$$

and form a chain rule to describe the ODE in terms of $$\displaystyle u$$ and $$\displaystyle \dot{u}$$. That ODE will then be much easier to solve.

Have a go Give us a shout if you get stuck.

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#### luimucar

I reviewed how to solve 1st order EDO. In this case, it seems that original EDO is separable, so I do not understand why it is necessary to do that substitution.

However, I am confused because solving the EDO by the separable method, returns a complex expression.

So I tried to follow your substitution but I am no able to separate terms and solve it.

Could u please go on with your expressions to reach the solution?

Thanks

#### benit13

I reviewed how to solve 1st order EDO. In this case, it seems that original EDO is separable, so I do not understand why it is necessary to do that substitution.
However, I am confused because solving the EDO by the separable method, returns a complex expression.
Ahh yes, the equation is separable, so my additional substitution is overkill. It's just a force of habit for me to substitute like that because I hate dealing with square roots in ODEs.

I don't know how you ended up with something complex though...

$$\displaystyle \frac{d\phi}{dt} = A \phi^{1/2}$$

$$\displaystyle \int \phi^{-1/2} d \phi = \int A dt$$

$$\displaystyle 2 \phi^{1/2} = At + c$$

$$\displaystyle \phi = \frac{1}{2}(At+c)^2$$

Then differentiate that twice to get $$\displaystyle \ddot{\phi}$$

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