- #1

Jeavenasti

- 5

- 2

- Homework Statement
- A rod with mass M (without a fixed axis) is attached to two masses m (they are point masses) located at its ends. The system rotates about the center of mass with velocity ##w_o## k . At the instant in which a mass is in the position L/2 i, it detaches from the rod leaving with the same instantaneous speed that it had at that moment. Calculate all the new velocities.

- Relevant Equations
- Angular momentum = I x w + r x mv

First, I can say that the velocity of the mass leaving the system is equal to ##w_o## (k) x ##L/2## (i) which tells me that its velocity will be w_o L/2 (j)

Now, since the net external force is equal to 0, (linear) momentum is conserved, so:

At first the velocity of the center of mass was 0 and after it leaves the system, it will still be 0, so I can state:

0 = mv_1 + 2mv_2

(being v_1 = velocity of the mass leaving the rod and v_2 the velocity of the rod and the other mass together).

So:

v_2 = -mv_1/2m

v_2 $= - v_1/2 = -w_o L/4 (j)----------Now, I need to find the angular velocities:

since the sum of torques is 0, angular momentum is conserved (assume it is true).

So,

L_o = L_f

I can calculate L_o using my Center-of-momentum frame in the center of mass of the system mass-rod-mass.

L_o = I w = (1/12 mL^2 + 2 m(L/2)^2)w_o = 7/12 mL^2

Now, here I had a few doubts.

First of all, I found that the rod-mass system (with only 1 mass) will have a new center of mass in i-axis equals to $x_{cm1} = -L/4$ i (this comes from, xcm = m (-L/2)/2m, with (0,0) in the center of the rod)

But then, to find the final Angular Momentum I had doubts:

Do I have to change the moment of inertia of the rod-mass with respect to an axis that passes through the center of mass of the rod-mass system? Something like:

I_f = (1/12 mL^2 + m (L/4)^2) + m(3/4 L)^2 = 17/24 m L^2

(Or Do I have to keep the same moment of inertia?).

Then, I can write:

L_f = 17/24 m L^2 w_f + (L/2) (i) x m v_1 (j) + (-L/4) (i) x 2m v_2 (-j)

L_f = 17/24 m L^2 w_f + (L/2) m v_1 (k) + (L/4) 2m v_2 (k)

But, since the modules of v_1 and v_2 are: ##v_1 = w_oL/2##, and ##v_2 = w_oL/4##

L_f = (17/24) m L^2 w_f + (L/2) m (w_oL/2) (k)+ (L/4) 2m (w_oL/4) (k)

L_f= 17/24 m L^2 w_f + (Lm w_oL)/4 (k)+ (Lm w_oL)/8 (k)And then, setting it equal to the initial angular momentum, I can determine wf. My main doubt arises from what I mentioned above, I don't know if changing the moment of inertia of the rod-mass is correct (and I also have doubts, could I have placed the center of moments at another point to do something easier?). I´d apreciatte some help, thanks in advance. (Note: I have not solved it rigorously and it is likely that some things are not well justified, but I shared it to solve specific doubts)

Now, since the net external force is equal to 0, (linear) momentum is conserved, so:

At first the velocity of the center of mass was 0 and after it leaves the system, it will still be 0, so I can state:

0 = mv_1 + 2mv_2

(being v_1 = velocity of the mass leaving the rod and v_2 the velocity of the rod and the other mass together).

So:

v_2 = -mv_1/2m

v_2 $= - v_1/2 = -w_o L/4 (j)----------Now, I need to find the angular velocities:

since the sum of torques is 0, angular momentum is conserved (assume it is true).

So,

L_o = L_f

I can calculate L_o using my Center-of-momentum frame in the center of mass of the system mass-rod-mass.

L_o = I w = (1/12 mL^2 + 2 m(L/2)^2)w_o = 7/12 mL^2

Now, here I had a few doubts.

First of all, I found that the rod-mass system (with only 1 mass) will have a new center of mass in i-axis equals to $x_{cm1} = -L/4$ i (this comes from, xcm = m (-L/2)/2m, with (0,0) in the center of the rod)

But then, to find the final Angular Momentum I had doubts:

Do I have to change the moment of inertia of the rod-mass with respect to an axis that passes through the center of mass of the rod-mass system? Something like:

I_f = (1/12 mL^2 + m (L/4)^2) + m(3/4 L)^2 = 17/24 m L^2

(Or Do I have to keep the same moment of inertia?).

Then, I can write:

L_f = 17/24 m L^2 w_f + (L/2) (i) x m v_1 (j) + (-L/4) (i) x 2m v_2 (-j)

L_f = 17/24 m L^2 w_f + (L/2) m v_1 (k) + (L/4) 2m v_2 (k)

But, since the modules of v_1 and v_2 are: ##v_1 = w_oL/2##, and ##v_2 = w_oL/4##

L_f = (17/24) m L^2 w_f + (L/2) m (w_oL/2) (k)+ (L/4) 2m (w_oL/4) (k)

L_f= 17/24 m L^2 w_f + (Lm w_oL)/4 (k)+ (Lm w_oL)/8 (k)And then, setting it equal to the initial angular momentum, I can determine wf. My main doubt arises from what I mentioned above, I don't know if changing the moment of inertia of the rod-mass is correct (and I also have doubts, could I have placed the center of moments at another point to do something easier?). I´d apreciatte some help, thanks in advance. (Note: I have not solved it rigorously and it is likely that some things are not well justified, but I shared it to solve specific doubts)

Last edited by a moderator: