Calculating Moment of Inertia and Angular Motion of Stick

[Naeem]

A stick of uniform density with mass M = 7.6 kg and length L = 0.8 m is pivoted about an axle which is perpendicular to its length and located 0.13 m from one end. Ignore any friction between the stick and the axle.


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a) What is the moment of inertia of the stick about this axle?
Iaxle = kg m2
1.621333 NO

HELP: Use the parallel-axis theorem.

i.e Icom + Mh2, how to apply that here.

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The stick is held horizontal and then released.
b) What is its angular speed as it passes through the vertical

w = rad/s


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c) What is its angular acceleration as it passes through the vertical position?
a = rad/s2


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d) What is the magnitude of the vertical component of the force exerted by the stick on the axle when the stick passes through the vertical?
|Fvertical| = N


Somebody help me with all these parts!

 

[jmerickson]

did you figure out part d yet i am working on the same problem

 

[wais]

a) Using the parallel-axis theorem, we can calculate the moment of inertia of the stick about the axle as:
Iaxle = Icom + Mh^2
Where Icom is the moment of inertia of the stick about its center of mass, M is the mass of the stick, and h is the distance between the center of mass and the axis of rotation (in this case, h = 0.13 m).
We can calculate Icom for a thin rod rotating about its center of mass using the formula:
Icom = (1/12)ML^2
Plugging in the given values, we get:
Icom = (1/12)(7.6 kg)(0.8 m)^2 = 0.512 kg m^2
Therefore, the moment of inertia of the stick about the axle is:
Iaxle = 0.512 kg m^2 + (7.6 kg)(0.13 m)^2 = 1.621333 kg m^2

b) When the stick passes through the vertical position, it will have converted all of its potential energy into kinetic energy. Using the conservation of energy, we can calculate the angular speed as:
1/2Iw^2 = mgh
Where I is the moment of inertia, w is the angular speed, m is the mass, g is the acceleration due to gravity, and h is the height of the stick's center of mass.
Since the stick's center of mass starts at a height of 0.8 m and ends at a height of 0.13 m, we can calculate the height as:
h = (0.8 m + 0.13 m)/2 = 0.465 m
Plugging in the values, we get:
1/2(1.621333 kg m^2)(w)^2 = (7.6 kg)(9.8 m/s^2)(0.465 m)
Solving for w, we get:
w = √((2(7.6 kg)(9.8 m/s^2)(0.465 m))/(1.621333 kg m^2)) = 5.424 rad/s

c) Since the stick is rotating at a constant speed as it passes through the vertical, its angular acceleration is 0.

d) At the vertical position, the stick is exerting