Angular Acceleration of A rod with friction. Dynamics (General Plane)

[viperking]

Homework Statement

A 100 kg slender rod is lifted on the left end until the angle between the ground and the right end is 30 degrees. The right end is still in contact with the ground. The coefficient of friction between the rod and the ground is .5 and the length of the rod is 2.1 m.

Homework Equations

∑fx=m(ag)x
∑fy=m(ag)y
∑M_g=I_g∂
F_f=.5Fn

The Attempt at a Solution

I summed the forces in the x direction and got -F_f=m(ag)x
In the y direction I got -981+F_n=m(ag)y but i think the y acceleration of the center of gravity may be zero because of the fact that the bar can't translate up or down due to the contact with the ground. It can only rotate or slip horizontally on the ground. Finally i summed the moments about the center of gravity and got that fn(1.05cos(30))+Ff(1.05sin(30))=36.75∂

I got a final answer of 17.26 rad/s^2 counter clockwise, but i don't think it is correct. Any suggestions?

 

[paisiello2]

What is the actual question?

 

[viperking]

What is the actual question?

Find the angular acceleration at the instant it is released. So the angle is 30 degrees

 

[paisiello2]

The center of mass will accelerate in the y-direction as well as the x-direction.

 

[rezeew]

I would suggest double checking your calculations and assumptions. It is important to consider all forces acting on the rod, including the normal force and the force of friction. Additionally, the angle of the rod and the coefficient of friction may also affect the calculations. It may also be helpful to draw a free body diagram to visually understand the forces at play. If you are still unsure, I would recommend consulting with a colleague or using a physics simulation program to verify your answer.