Solve Rotational Problem 3: Max Angle for Rod+Clay

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The discussion revolves around a physics problem involving a rod and a ball of clay, focusing on the conservation of angular momentum and energy. Participants clarify that angular momentum is conserved only about specific points where external forces do not contribute. The kinetic energy of the clay and rod system is analyzed, noting that after the clay sticks to the rod, the system's energy shifts from kinetic to potential as it rotates. It is emphasized that linear momentum is not conserved during the impact due to external forces acting on the system. The conversation highlights the importance of correctly identifying the moment of inertia and the reference point for calculating angular momentum.
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Homework Statement



A 75 g 30 cm long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 10g ball of clay traveling horizontally at 2.5 m/s hits and stick to the very bottom tip of the rod. To what maximum angle measured from the vertical, does the rod with attached ball of clay rotate?

Homework Equations


Angular momentum is conserved
initial L=final L
mvr=Iω
I=1/12MR^2 +mr^2
rotational K.E=1/2 Iω^2
MGh

The Attempt at a Solution


angular momentum :
0.01(2.5)(0.3)=(I1+I2)ω
Rod: I1=1/12MR^2=1/12(0.075)(0.15)^2=1.40e-04
Clay: I2=mR^2=.01(0.15)^2=2.25e-4
Initial energy =1/2 (I1+I2)ω^2

h1=.3-.3cos∅
mgh1=.01(9.8)(.3-.3cos∅)
h2=.15-.15cos∅
Mgh2=0.075(9.8)(.15-.15cos∅)
Final energy=Mgh1+mgh2
my question is only kinetic rotational energy is converted to potential energy and how come the kinetic energy is not converted to rotational energy and my teacher said the linear momentum is not conserved and I am so confused . Secondly, why the mass of initial angular momentum is small mass not the mass of clay and rod ?
 
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Anthonyphy2013 said:
Angular momentum is conserved
Only if you choose the point to take moments about such that external forces make no contribution.
angular momentum :
0.01(2.5)(0.3)=(I1+I2)ω
About what point? How far is the point of impact of the clay ball from that point?
Rod: I1=1/12MR^2=1/12(0.075)(0.15)^2=1.40e-04
It's 1/12 if R is the length of the whole rod. What length did you use above?
h1=.3-.3cos∅
What does h1 represent?
my question is only kinetic rotational energy is converted to potential energy
You can view the clay mass after impact either as rotating about the centre of the rod (so use rotational KE) or as moving linearly horizontally (briefly). They produce the same KE. These are two different ways of looking at the same thing - don't add them.
my teacher said the linear momentum is not conserved
That's right. It obviously is not conserved during impact, and it's not conserved during the subsequent rotation either because there will be a force from the axle.
Secondly, why the mass of initial angular momentum is small mass not the mass of clay and rod ?
Initial? Do you mean before or after impact? Before impact, the rod is stationary, so how would that contribute to angular momentum?
 
I have another question if the massless rod is in that case , should I consider the Kinetic energy to be the initial energy ?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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