# Conservation of Angular momentum problem

• i_hate_math
In summary, the small block slides down a frictionless surface through height h=0.471 m and then sticks to a uniform vertical rod of mass M=0.470 kg and length d=2.36 m. The rod pivots about point O through angle θ before momentarily stopping. Find θ.f
is h the original height and l the final height?

How did you get l=h?

How did you get l=h?
L=(0.5I*w^2)/mg

Hmm... I don't think energy is conserved as it is an inelastic collision.

Hmm... I don't think energy is conserved as it is an inelastic collision.
After the collison, mechanical energy is conserved

now w=sqrt(2gh)/d
mgl=0.5Iw^2, where l is change in height
Right, though we will have to be careful not to confuse l (lower case L,)with I (uppercase I). To avoid confusion I'll write H for the change in height and Imom for moment of inertia. Also, the m in there should be for the mass of the rod plus block, not just the block, and H is the change in height of the mass centre of that combination. We will return to that later.

As I said, you next need to figure out the right value for Imom. This is for the rod and block as a combined system. Think carefully.

Right, though we will have to be careful not to confuse l (lower case L,)with I (uppercase I). To avoid confusion I'll write H for the change in height and Imom for moment of inertia. Also, the m in there should be for the mass of the rod plus block, not just the block, and H is the change in height of the mass centre of that combination. We will return to that later.

As I said, you next need to figure out the right value for Imom. This is for the rod and block as a combined system. Think carefully.
Since the block is treated as a particle, the Imom would be 1/3mw^2 right?

Since the block is treated as a particle, the Imom would be 1/3mw^2 right?
(1/3)(m+M)w^2 is what i meant.

Let ω be the instantaneous angular velocity of the rod as well as the stuck block about the pivot just after impact.
Te expression for angular momentum of rod + block just after impact {(Md²/3) + (md²)}ω. Equating this with initial angular momentum of the block just before impact gives you ω
To find θ, find the change in potential energy of the rod and the stuck block, when the rod reaches the angle θ and momentarily come to rest. This then is ro be equated with initial rotational KE of the of the block using formula 0.5Iω², where I = {(M/3)+m}d². { Note I(rod) = M(d²/3) and that of stuck small block = md²}

i_hate_math
(1/3)(m+M)w^2 is what i meant.
First, you don't mean w2, I hope. A moment of inertia is a mass multiplied by the square of a distance.
Secondly, you need to consider each body separately and add up their moments. What did you find for the moment of the block about O in post #26?

Let ω be the instantaneous angular velocity of the rod as well as the stuck block about the pivot just after impact.
Te expression for angular momentum of rod + block just after impact {(Md²/3) + (md²)}ω. Equating this with initial angular momentum of the block just before impact gives you ω
To find θ, find the change in potential energy of the rod and the stuck block, when the rod reaches the angle θ and momentarily come to rest. This then is ro be equated with initial rotational KE of the of the block using formula 0.5Iω², where I = {(M/3)+m}d²
Please do not spoon-feed @i_hate_math. He/she needs to get the hang of figuring these things out.

First, you don't mean w2, I hope. A moment of inertia is a mass multiplied by the square of a distance.
Secondly, you need to consider each body separately and add up their moments. What did you find for the moment of the block about O in post #26?
Yeah i didnt. I think i know where i went wrong. Its MoI again, nasty little bugger, the MoI of a point is simply md^2, and i forgot to add this into my calculation.