Separating angular and linear momentum

[simbil]

Do I need to add more of my working to make it clearer to follow?

 

[Doc Al]

Finally got around to this again and I have worked out what may be a solution.

So, following your advice to find the velocity of the centre of mass as a function of angle:

We know that the energy E is constant such that the gravitic potential and kinetic are constant so E = mgh - 0.5mv^2

Before the rod falls, v is zero and h of COM = 1m, unit mass, so,

E = g

Therefore, g = mgh - 0.5mv^2
g = gh - 0.5v^2

Height can be expressed as a function of angle as h = rcosθ = cosθ, so,

g = gcosθ - 0.5v^2

and so the velocity as a function of angle is,

v^2 = 2g - 2gcosθ

If I am correct so far, the next step is to calculate the radial acceleration = v^2/r = v^2, so,

radial a = 2g - 2gcosθ

Tangential acceleration has already been calculated for the example as 0.75gsinθ

Looks good.

Total acceleration horizontally = tangential.cosθ - radial.cosθ

Almost. The horizontal component of the radial acceleration would be -radial.sinθ.

a = 0.75gsinθcosθ - cosθ(2g - 2gcosθ)

Unit mass so applying F= ma, horizontal force = 0.75gsinθcosθ - cosθ(2g - 2gcosθ)

Putting numbers into there gives results in line with what I would expect - is it right..?

Correct the above and you've got it.

Do I need to add more of my working to make it clearer to follow?

Nope. Just took me a while to get around to it.

 

[simbil]

Thanks Doc Al, appreciate your guidance.



horizontal force = 0.75gsinθcosθ - sinθ(2g - 2gcosθ)