Hinged rod rotating, falling and hitting a mass

[Very_Unwise]

Homework Statement

A thin rod with length L and mass M pivoted at one end falls (rotates) frictionlessly from a position parallell to the horisontal, starting from rest. At the moment the rod is completely vertical, it collides with a snowball laying on ice with mass=1/9 M. The snowball sticks on to the rod at the tip of the rod. What is the angular velocity of the combined object (rod+snowball) immediatly after collision?

Relevant Equations

Ep = M*g*h
I = 1/3 * M*L^2
Ek= 1/2*I *w^2
P = m*r*v

Assuming no friction anywhere, no drag and perfect inelastic collision

Using conservation of mechanical energy i can determine the rotational speed of the rod right before collision occurs.
mgh=1/2*i*w^2

center of mass falls 1/2*L so we have:
M*g*1/2*L = 1/2*(1/3*M*L^2)*w^2

Solving for w gives w^2=(3*g)/L , g=9.81
w=sqrt (29.43/L)
Using conservation of angular momentum P before = P after collision
m*r*v = m*r*v , v=w*r , r=L
M*w before collision *L^2 = (M+1/9*M)*w*L^2 after collision, divide by L^2

M*sqrt (29.43/L) = 10/9*M *w after collision
solving for w after collision gives

w= 4.88/L
Is this the correct way to do it? I don't think i can use conservation of rotational kinetic energy as the collision is inelastic, not elastic.
I also have P =I * w but i don't see a use for it

 

[Doc Al]

Is this the correct way to do it?

Partly. I didn't check your calculations, but you had the correct approach for finding the rotational speed just prior to the collision.

I don't think i can use conservation of rotational kinetic energy as the collision is inelastic, not elastic.

That's right. Mechanical energy is not conserved during that collision.

I also have P =I * w but i don't see a use for it

That's where you messed up. You need to use that to express the angular momentum before and after the collision. For some reason, you used m*r*v.

 

[Very_Unwise]

Okay, so instead of m*r*v=m*r*v ill use I*ω = m*r*v?
Substituting v=ω*L gives me I*ω = m*L^2*ω

That gives me i*ω = 10/9*M *L*v v=ω*L
i*ω1 =10/9*M*L^2 * ω2 , i=1/3*M*L^2, ω1 = sqrt (29.43/L), I=Irod
I rod*sqrt (29.43/L) = 10/9 *M*L^2*w

solving for w using symbolab gives me w = (1.627*sqrt (L))/L

I also expect a pretty small change in angular velocity as the weight of the snowball is quite small.
If i give L a value of 10 i get w (before) =sqrt(29.43/10) which is 0.542 rad/s
w(after) = (1.627*sqrt(10)/10 which is 0.5145 rad/s. The answer seems plausible atleast

 

[Doc Al]

Okay, so instead of m*r*v=m*r*v ill use I*w = m*r*v? I am thinking that i have to use m*r*v somewhere as the mass of the snowball is stated in the homework statement.

You'll certainly need to include the mass of the snowball in your analysis. You'll want to express everything in terms of angular velocity ω, not v. Before the collision, only the rod is moving. And you know the moment of inertia of that rod (call it I_rod). What's the moment of inertia of the rod + snowball system?

What's the moment of inertia of the snowball? Hint: You had the right idea with m*r*v, but express that in terms of ω.

Before the collision, call the angular velocity ω₁; after the collision, call the angular velocity ω₂. You found ω₁. Now set up an equation to solve for ω₂.

 

[Very_Unwise]

Okay, so my previous calculations were wrong because the moment of inertia for the rod changes after collision? Meaning I(combined object) doesn't equal Irod? To find the new moment of inertia i have to find the new center of mass for the combined object, which is (0*1/9*M +1/2*L*M)/10/9*m.

That gives us r = 11/20*L (center of mass moves "down" towards the end of the rod with the snowball attached) Moment of inertia of thin rod is 1/12*M*L^2. The distance to the axis of rotation for the rod+snowball would be 11/20*L
The new moment of inertia would be I + M*d^2 which is is 1/12*M*L^2 + M*(11/20*L)^2.
That gives me I = (463*M*L^2)/1200, where M is the combined weight

As for the moment of inertia for the snowball that's 2/5*M*L^2 + M*L^2 which gives me
I = 7/5*M*L^2 where M is the weight of the snowball.

I don't see where i could use this though as the snowball doesn't move before collision? Do i use this afterwards? So as example for the equation of conservation of angular momentum would be something like:

Irod* ω1= I(rod+snowball)*ω2 + I(snowball) *ω2?
Or just Irod* ω1= I(rod+snowball)*ω2?

Edit: my calculations may be incorrect but I am more worried about if I am thinking correctly? Or am i way off here?

 

[haruspex]

w=sqrt (29.43/L)

Maybe you have been taught otherwise, but in a problem where all the data are provided symbolically, not as numbers, I would leave g as symbolic too. $\sqrt{\frac{3g}L}$ will serve as the answer.
If you are going to substitute a numeric value then you should include the units: $\sqrt{\frac{29.43}{L}}m^{\frac 12}s^{-1}$, which is rather horrible, but necessary.

 

[Very_Unwise]

Maybe you have been taught otherwise, but in a problem where all the data are provided symbolically, not as numbers, I would leave g as symbolic too. $\sqrt{\frac{3g}L}$ will serve as the answer.
If you are going to substitute a numeric value then you should include the units: $\sqrt{\frac{29.43}{L}}m^{\frac 12}s^{-1}$, which is rather horrible, but necessary.

Yes that's true, substituting g=9.81 does make things quite messy. Thank you

 

[Doc Al]

Okay, so my previous calculations were wrong because the moment of inertia for the rod changes after collision? Meaning I(combined object) doesn't equal Irod?

The moment of inertia of the "rod + snowball" is just the sum of the moments of inertia of each. What might be new to you is the idea that a point mass (the snowball) has a moment of inertia about the pivot. Check this out: Moment of Inertia

To find the new moment of inertia i have to find the new center of mass for the combined object, which is (0*1/9*M +1/2*L*M)/10/9*m.

Nope. No need to find the new center of mass. I_{combined object} = I_rod + I_snowball

Or just Irod* ω1= I(rod+snowball)*ω2?

Yes!

 

[Very_Unwise]

Okay, so I(rod+snowball) should be Irod + Isnowball which is
1/3*M*L^2+ m*L^2 , m=1/9*M
I(rod+snowall) = 1/3*M*L^2 + 1/9*M*L^2 = 4/9*M*L^2

Irod* ω1= I(rod+snowball)*ω2
1/3*M*L^2 * sqrt(3g/L) = 4/9*M*L^2*ω2
crossing out M*L^2
1/3*sqrt(3g/L)=4/9*ω2
ω2*4/3=sqrt(3g/L)
ω2=sqrt(3g/L)*3/4

 

[haruspex]

Okay, so I(rod+snowball) should be Irod + Isnowball which is
1/3*M*L^2+ m*L^2 , m=1/9*M
I(rod+snowall) = 1/3*M*L^2 + 1/9*M*L^2 = 4/9*M*L^2

Irod* ω1= I(rod+snowball)*ω2
1/3*M*L^2 * sqrt(3g/L) = 4/9*M*L^2*ω2
crossing out M*L^2
1/3*sqrt(3g/L)=4/9*ω2
ω2*4/3=sqrt(3g/L)
ω2=sqrt(3g/L)*3/4

Good.
Notice how the 1/9 mass has a disproportionate effect because of its increased distance from the hinge.

 

[Doc Al]

Okay, so I(rod+snowball) should be Irod + Isnowball which is
1/3*M*L^2+ m*L^2 , m=1/9*M
I(rod+snowall) = 1/3*M*L^2 + 1/9*M*L^2 = 4/9*M*L^2

Irod* ω1= I(rod+snowball)*ω2
1/3*M*L^2 * sqrt(3g/L) = 4/9*M*L^2*ω2
crossing out M*L^2
1/3*sqrt(3g/L)=4/9*ω2
ω2*4/3=sqrt(3g/L)
ω2=sqrt(3g/L)*3/4

Now you've got it. Good work!

 

[Very_Unwise]

Alright alright, thanks for the help :D