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Rigid body kinetics

  • Thread starter 9988776655
  • Start date
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1. Homework Statement

The problem is shown in the attachment.

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

3. 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:

learningphysics

Homework Helper
4,099
5
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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