- #1
Telemachus
- 835
- 30
Well, it's my second post about rigid body. I originally posted this on introductory physics, but as nobody answered this, or the previous topic, I've decided to post this here.
I have this other exercise rigid in the plane, with which I am having problems.
The rod of mass m and length l, is released based on the vertical position of rest with the small roller end A resting on the slope. Determine the initial acceleration A.
(neglect friction and mass of the roller A)
The answer given by the book is [tex]a_A=\displaystyle\frac{g\sin \theta}{1-\frac{3}{4}\cos^2\theta}[/tex]
I try to raise the moment equation, and Newton. But not me, not that I'm doing wrong. For this consider that the bar rotates about its center of mass.
[tex]I_{cm}=\displaystyle\frac{mL^2}{12}[/tex]
[tex]N-mg\cos\theta=0[/tex]
[tex]mg\sin\theta=ma_{cm}[/tex]
[tex]I_{cm}\alpha=\displaystyle\frac{L}{2}mg\sin\theta\cos\theta[/tex]
[tex]\alpha=\displaystyle\frac{6g\sin\theta\cos\theta}{L}[/tex]
[tex]a_cm=g\sin\theta[/tex]
Then: [tex]a_A=a_{cm}+\displaystyle\frac{L}{2}\alpha=g\sin\theta+3g\cos\theta\sin\theta[/tex]
Greetings and thanks for posting.
Homework Statement
I have this other exercise rigid in the plane, with which I am having problems.
The rod of mass m and length l, is released based on the vertical position of rest with the small roller end A resting on the slope. Determine the initial acceleration A.
(neglect friction and mass of the roller A)
The answer given by the book is [tex]a_A=\displaystyle\frac{g\sin \theta}{1-\frac{3}{4}\cos^2\theta}[/tex]
Homework Equations
I try to raise the moment equation, and Newton. But not me, not that I'm doing wrong. For this consider that the bar rotates about its center of mass.
[tex]I_{cm}=\displaystyle\frac{mL^2}{12}[/tex]
[tex]N-mg\cos\theta=0[/tex]
[tex]mg\sin\theta=ma_{cm}[/tex]
[tex]I_{cm}\alpha=\displaystyle\frac{L}{2}mg\sin\theta\cos\theta[/tex]
The Attempt at a Solution
[tex]\alpha=\displaystyle\frac{6g\sin\theta\cos\theta}{L}[/tex]
[tex]a_cm=g\sin\theta[/tex]
Then: [tex]a_A=a_{cm}+\displaystyle\frac{L}{2}\alpha=g\sin\theta+3g\cos\theta\sin\theta[/tex]
Greetings and thanks for posting.