Rigid Body Kinetics Homework Solution

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SUMMARY

The discussion focuses on solving a rigid body kinetics problem involving a bar and a pulley system. The moment of inertia for the bar is corrected to (1/3)ML², contrasting with the incorrect assumption of MgL/2. Key equations include torque equations for the rod and force equations for the hanging mass, which are essential for determining angular acceleration and tension. The analysis emphasizes calculating both vertical and horizontal accelerations of the center of mass to derive the forces acting on the system.

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Homework Statement



The problem is shown in the attachment.

Homework Equations


Mo = Ia
MgL/2 = moment of inertia of the bar
Io = Ig + md^2 parallel axis theorum

The Attempt at a Solution



I found the moment of inertia of the bar about its center of mass:
MgL/2 = (4.6*9.81*1.3)/2 = 29.3319
For the pulley, I tried:
(T2-T1)radius = Ialpha
acceleration = alpha*radius
For the mass:
T-mg = ma
There are too many unknowns for me unfortunately
 
Last edited:
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There's probably multiple ways to do this problem... but the way that comes to mind for me is... first get the tension in the cable and get the angular acceleration of the rod about the pivot...

The moment of inertial of the bar is not MgL/2... The moment of inertia of a rod about it's end is (1/3)ML^2.

Use your torque equation for the rod... with the force equation for the hanging mass... to get alpha and tension...

Once you do that, you focus on the force equations for the rod...

What is the vertical acceleration of the center of mass of the rod? you can get this using your alpha...

What is the horizontal acceleration of the center of mass of the rod? you can get this using the v given for the mass (from this v you can get w, the angular velocity of the rod... then you can get the vertical velocity of the center of mass)...

from the vertical velocity of the center of mass, you can get the horizontal acceleration... think centripetal motion...

with the horizontal and vertical accelerations of the center of mass... you can solve for the forces at the A...

sum of all horizontal forces acting on the rod = (mass of the rod)*(acceleration of the center of mass of the rod)...

same thing vertically.
 

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