How to find replacement function of a mass?

Question

I wonder how I can find the replacement function of the center of the blue mass? The center of mass of the blue mass is $(0,0)$ and the blue mass is homogeneous. The masses do not move at t=0 in the system (Their initial speeds are zero). There is no friction in the system and the red mass is a point mass. $u$ is the angle of the red mass with $y$ axis.

I need help in learning the theory behind solving such systems. Thank you.

EDITED: I would like to show what I did till now to solve the problem. I followed the hints that Manishearth offered

$u=\alpha _0$ The definations of the blue mass center replacement:

$X_b(t)$ The replacement function that I try to fnd

$Y_b(t)$

The definations of the red mass Center replacement:

$X_r(t)$

$Y_r(t)$

1- Mass of center of entire system is in the beginning: $$r_{0x}=\frac{m. r. sin(\alpha _0)}{m+M}$$ $$r_{0y}=-\frac{m. r. cos(\alpha _0)}{m+M}$$ Mass center will not depend on time. If So,

$$mX_r(t)+MX_b(t)=(m+M)r_{0x}=m. r. sin(\alpha _0)$$ (1)

$$mY_r(t)+MY_b(t)=(m+M)r_{0y}=-m. r. cos(\alpha _0)$$ (2)

2-Energy is conserved

$$\frac{m (\dot X_r(t))^2 } {2}+ \frac{m (\dot Y_r(t))^2 } {2}+\frac{M (\dot X_b(t))^2 } {2}+\frac{M (\dot Y_b(t))^2 } {2} +mg(a+Y_r(t))+Mg(a+Y_b(t))=mg(a-r.cos(\alpha _0))+Mga$$ (3)

I dont know how to apply Hint 3 and Hint 4. Till now I have 4 unknown and 3 equations. Am I in correct way to solve $X_b(t)$ ? Thanks in advice

Answer 1

I'm not sure what your level of understanding of physics is, so I've broken it up into hints. If you do not understand the meaning of a hint, ask me and I'll elaborate.

Hint 1: The center of mass of the entire system should stay constant in the x-direction. This will give you a relation between the x-coordinates of the two objects. This can be differentiated to get relations between object velocities and accelerations (in x direction)

Hint 2: Energy is conserved as well.

Hint 3: Use relative velocities/accelerations; work in the frame of the blue mass.

Hint 4: There is circular motion going on in the frame of the blue body. Note that we can only apply our usual $m\omega^2 r$ formula in the moving frame of the body.