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TerpFlacco
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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.
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
p=mv
L=Iω
Erot = .5*I*ω2
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...
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...
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