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Angular acceleration of a rod?

  • Thread starter euphtone06
  • Start date
  • #1
22
0

Homework Statement


http://img505.imageshack.us/img505/6741/rodprobbw3.gif" [Broken]

Homework Equations


Itot= 1/3MrL^2 + 2/5 MsR^2 + Ms(L + R)^2 = 235.69784
Cm= (Mr(L/2) + Ms(L+r))/(Mr+Ms) = 4.915

angular accel = Torque/Itot
Torque= 1/2Lmgcos(theta)

The Attempt at a Solution


I solved for Itot and Center of mass and got the correct answers but, I am having a difficult time getting the angular acceleration correct.

Aa= .5Lmgcos(theta)/(Itot)
= .5*6*4*9.8cos31/235.697
= .42767 rad/s^2 which was incorrect
Please help dont know what went wrong!
 
Last edited by a moderator:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
The 'L' in your angular acceleration formula should be the distance from the axis to the point where the force causing the torque is exerted. In this case the gravitational force can be treated as being exerted at the center of mass, not at the end of the rod.
 
  • #3
22
0
Do you have to account for the total mass of the system?
.5*4.915*(4+4???)*9.8*cos31/235.697= .7 rad/s^2
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
4+4??? Don't you mean 4+1? Yes, total mass of system. And it looks like the 1/2 in your torque equation is coming from the assumption that the center of mass is the center of mass of the rod. Get rid of it and just put L to be distance from axis to center of mass.
 

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