Max Angle Rod Rotates when Hit w/ Ball

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SUMMARY

The maximum angle at which a 40-cm-long rod rotates when struck by a 14 g ball of clay traveling at 2.4 m/s is determined through the conservation of energy principles. The momentum of the ball is calculated as 0.0336 kg-m/s, leading to an angular momentum of 0.01344 kg-m^2/s. The moment of inertia (I) of the system is computed as 0.0064 kg-m^2, resulting in an angular velocity (ω) of 2.1 rad/s. The rotational energy is found to be 0.014112 J, which is converted into potential energy to find the maximum angle from vertical.

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EDIT: Title should say "To what maximum angle, measured from vertical, does the rod rotate when hit will a ball?" Did not see that it had cut off. Sorry.

Homework Statement


A 78 g, 40-cm-long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 14 g ball of clay traveling horizontally at 2.4 m/s hits and sticks to the very bottom tip of the rod.

To what maximum angle, measured from vertical, does the rod (with the attached ball of clay) rotate?

MR = .078kg
L=.4m
MB = .014kg
v= 2.4m/s

Homework Equations


p=mv
L=Iω
Erot = .5*I*ω2

The Attempt at a Solution


I start off by finding the momentum of the ball as MB*v = 0.0336 kg-m/s

Then, I find angular momentum as 0.01344 kg-m^2/s

Next, I use I = (MR/3 + MB)*L^2 to get I as 0.0064 kg-m^2

Using angular momentum equaling I*omega, I find omega as 2.1 rad/s.

To find rotational energy, I used .5*I*omega and got 0.014112J

And now I do not know what to do...
 
Last edited:
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You can use angular speed, but you don't need to.
What you just need is to keep in mind that the kinetic energy of the ball will translate into potential energy, that is mgh=(mv^2)/2.
The rod does not contribute to this energy transfer as it's axle passes through its center.
 

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