 #1
jfnn
Homework Statement
A thin uniform rod (of mass 10.0 Kg and length of 1.20 m) is attached to a frictionfree pivot. Initially, the rod is balanced vertically above the pivot (position A in the figure attached). If the rod falls from rest, calculate
a. the angular acceleration at position B
b. the angular velocity at position C
Homework Equations
[/B]
m= 10.0 kg
l= 1.20 m
I=ml^2 / 3
v initial = 0 m/2
Torque (T) = moment of inertia (I) * angular acceleration (a)
The Attempt at a Solution
a. [/B]angular acceleration = T / I
where I is known > a = T / (ml^2 / 3) we know mass and l, therefore plug in to obtain:
a = T / 4.8
Net torque only comes from the weight, thus it is: mg * 1/2*l * sin (90) > This is at position B
Net torque = (10)(9.8)(1/2)(1.2)sin(90)
Net torque = 58.8
a = 58.8/4.8 > 12.3 rad/s^2 clockwise (direction is given)
Is this part correct? It was the only way I could think about doing this...
b) I used energy to solve for the angular velocity at part c.
I know: mgh = 1/2Iw^2 (after eliminating variable/etc. did I eliminate all the correct things and have the correct variables remaining?)
w^2 = 2mgh/I where I is the moment of inertia, and was calculated to be 4.8
Thus,
w^2 = 2mgh/4.8 where m is known, g i known, and h is known
w^2 = 2* 10 *9.8*1.2 / 4.8
w^2 = 49
w = 7.00 rad/s clockwise (direction is given)
Is this correct? It is the only process I could think of how to solve...
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