# Determine the angular acceleration and angular velocity

jfnn

## Homework Statement

A thin uniform rod (of mass 10.0 Kg and length of 1.20 m) is attached to a friction-free 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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Mentor
Looks good to me!

• jfnn
jfnn
Looks good to me!
Thank you so much!