- #1

shmoop

## Homework Statement

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A uniform rigid rod with mass Mr = 2.7 kg, length L = 3.1 m rotates in the vertical xy plane about a frictionless pivot through its center. Two point-like particles m1 and m2, with masses m1 = 6.7 kg and m2 = 1.6 kg, are attached at the ends of the rod. What is the magnitude of the angular acceleration of the system when the rod makes an angle of 51.1 degrees with the horizontal? (m2 is 51.1 degrees above the horizontal, and m1 is 51.1 degrees below the horizontal).

## Homework Equations

Torque=Inertia*Angular Acceleration

Inertia of uniform rigid rod = mL^2/12

## The Attempt at a Solution

I began by calculating inertia, like this:

I = m

*L^2/12 = (2.7kg)(3.1m)2/12=2.16225 kg*m2

Then I calculated the gravitational forces from m1 and m2:

m1g=(6.7kg)(9.8m/s2)=65.66N

m2g=(1.6kg)(9.8m/s2)=15.68N

Then I determined the component of these forces which act perpendicular to the lever arm:

65.66N*cos51.1=41.23N

15.68N*cos51.1=9.84N

Then I calculated the net torque/moment (clockwise negative, counterclockwise positive):

Radius = 1/2(3.1m)=1.55m

Torque = (1.55m

*41.23N)-(1.55m*9.84N) = 48.65Nm

Then I calculated the angular acceleration:

Angular acceleration = Torque/Inertia

Angular acceleration = 48.65Nm/2.16kg*m2

Angular acceleration = 22.52 rad/s2

I have been told my answer is incorrect through an online system where I am able to check my answers, however I'm not sure where I went wrong. If anyone would be able to spot an error, or guide me in the right direction, it would be greatly appreciated!