Rigid body dynamics: A rod with two masses rotates down from a horizontal starting position

  • #1
MatinSAR
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Homework Statement
A rod of mass M and length L is rotating freely around an axis at one end. There are two particles attached to the rod at distances of L/3 and 2L/3 from the axis, with masses m and 2m, respectively. The system is only affected by gravitational forces.
A) Determine the angular acceleration of the rod and particles under these conditions. B) Calculate the angular velocity of the system when the rod is vertical after rotating 90 degrees.
Relevant Equations
Rigid body dynamics.
1717787688334.png

For the rod and the two particles ##x_1## is not zero but we have:$$x_2=x_3=0$$ We start with inertia tensor of the rod: $$ I_{ij} = \int \lambda ( \delta _{ij} \sum_{k} x^2_k -x_i x_j ) dx_1$$ $$ \lambda = constant = M/L $$ Using above equations, we have:
$$I_{11} = \dfrac M L \int ( x^2_2 +x^2_3 ) dx_1=0$$ $$I_{22} = \dfrac M L \int ( x^2_1 +x^2_3 ) dx_1=\dfrac {1}{3} ML^2$$ $$I_{33} = \dfrac M L \int ( x^2_1 +x^2_2 ) dx_1=\dfrac {1}{3} ML^2$$ $$I_{12}=...=I_{23}=...=0$$
For the two particles: $$I_{ij}=m_1(\delta _{ij} \sum_{k} x^2_{1k} -x_{1i} x_{1j})+m_2(\delta _{ij} \sum_{k} x^2_{2k} -x_{2i} x_{2j})$$ After Substituting from the information provided in problem, we get: $$I_{11}=m_1(x^2_{12}+x^2_{13})+m_2(x^2_{22}+x^2_{23}) = 0$$ $$I_{22}=m_1(x^2_{11}+x^2_{13})+m_2(x^2_{21}+x^2_{23}) =mL^2$$ $$I_{33}=m_1(x^2_{11}+x^2_{12})+m_2(x^2_{21}+x^2_{22}) =mL^2$$$$I_{12}=...=I_{23}=...=0$$
Therefore, the rigid body has an inertia tensor given by: $$I_{1}=0$$$$I_{2}=I_{3}=(\dfrac M 3 +m)L^2=\beta$$ Euler's equations becomes : $$ - \omega_1 \omega_2- \dot \omega_3=N_3 $$ $$\omega_1 \omega_3- \dot \omega_2=N_2$$ $$\dot \omega_1=N_1 $$
Could you please verify if the steps I have taken so far are correct? Any guidance or feedback you can provide would be very helpful.
 
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  • #2
I think that you are overcomplicating matters. This is a physical pendulum.

For part (A) find the distance ##L_{cm}## of the center of mass from the pivot, the moment of inertia ##I_{support}## about the support and plug in. For part (B) use energy conservation.
 
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  • #3
kuruman said:
I think that you are overcomplicating matters. This is a physical pendulum.

For part (A) find the distance ##L_{cm}## of the center of mass from the pivot, the moment of inertia ##I_{support}## about the support and plug in. For part (B) use energy conservation.
Are you suggesting that I consider the combined mass of ##M+3m## located at the center of mass of the rigid body, and then use the formula ##\alpha =## angular momentum / moment of inertia to calculate the angular acceleration?
By finding angular momentum using formula ##\dfrac {d \vec L}{dt}=N##.


and then use the formula ##\alpha =## N / moment of inertia ? (moment of inertia =##(M+3m)R^2_{cm}## )
 
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  • #4
MatinSAR said:
Are you suggesting that I consider the combined mass of ##M+3m## located at the center of mass of the rigid body, and then use the formula ##\alpha =## angular momentum / moment of inertia to calculate the angular acceleration?
By finding angular momentum using formula ##\dfrac {d \vec L}{dt}=N##.


and then use the formula ##\alpha =## N / moment of inertia ? (moment of inertia =##(M+3m)R^2_{cm}## )
Yes.
 
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  • #5
MatinSAR said:
consider the combined mass of M+3m located at the center of mass of the rigid body
That's rarely the easiest way. Better to calculate the moments of inertia separately then add them.
 
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  • #6
haruspex said:
Better to calculate the moments of inertia separately then add them.
Is this not what I have already done in first post?
kuruman said:
Yes.
Thank you for your help @kuruman , I'll try it.
 
  • #7
MatinSAR said:
Is this not what I have already done in first post?
I didn’t read that in detail because it was evidently much more complicated than it needed to be.
You should be able to write down the MoIs about the pivot straight off: ##ML^2/3+m(L/3)^2+2m(2L/3)^2##.
Similarly the torque about the pivot: ##(ML/2+mL/3+2m(2L/3))g##.
 
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  • #8
haruspex said:
I didn’t read that in detail because it was evidently much more complicated than it needed to be.
You should be able to write down the MoIs about the pivot straight off: ##ML^2/3+m(L/3)^2+2m(2L/3)^2##.
Similarly the torque about the pivot: ##(ML/2+mL/3+2m(2L/3))g##.
How can we determine that we only need the Moments of Inertia about the pivot when there are two other axes perpendicular to the pivot? I mean, Why MoI is 0 for those other 2 axes?

For example, if a sphere rotates about an axis passing through its center, moments of inertia about two perpendicular axes other than the rotation axis, exist.
 
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  • #9
MatinSAR said:
How can we determine that we only need the Moments of Inertia about the pivot when there are two other axes perpendicular to the pivot? I mean, Why MoI is 0 for those other 2 axes?

For example, if a sphere rotates about an axis passing through its center, moments of inertia about two perpendicular axes other than the rotation axis, exist.
This is a pendulum that rotates in a plane. The only direction that matters is the plane perpendicular because the angular acceleration is in that direction.
 
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  • #10
MatinSAR said:
How can we determine that we only need the Moments of Inertia about the pivot when there are two other axes perpendicular to the pivot? I mean, Why MoI is 0 for those other 2 axes?

For example, if a sphere rotates about an axis passing through its center, moments of inertia about two perpendicular axes other than the rotation axis, exist.
The two forces, gravity and the support force, define a plane.
The rod and masses are all in that plane.
 
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  • #11
Ok. I try again using the approach sugggusted by you.
$$\alpha = \dfrac { \sum_{i} r_i F_i}{I}=\dfrac {\frac {ML}{2}+\frac{mL}{3}+\frac{4mL}{3} }{(M/3 + m)L^2}g$$ And the angular acceleration is directed out of the plane.

haruspex said:
Similarly the torque about the pivot: ##(ML/2+mL/3+2m(2L/3))g##.
Didn't you forget ##\sin \theta##? If gravitional force acts downward, we have:
1717802380374.png



For part b, can I assume an arbitrary initial angular velocity ##\omega_0## and then use the conservation of energy to determine the angular velocity ##\omega## of the vertical rod?
I should just determine the change in height of the center of mass.
 
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  • #12
MatinSAR said:
For part b, can I assume an arbitrary initial angular velocity ##\omega_0## and then use the conservation of energy to determine the angular velocity ##\omega## of the vertical rod?
I should just determine the change in height of the center of mass.

CoM is located at: $$ R_{CoM} = \dfrac {m(L/3)+M(L/2)+2m(2L/3)}{M+3m}=\dfrac {(M/2) + (5m/3)}{M+3m}L$$So change in potential energy of the body is: $$\Delta U = (M+3m)(\dfrac {(M/2) + (5m/3)}{M+3m}L)g=(\dfrac M 2 + \dfrac {5m}{3})Lg $$ Using conservation of energy, we get: $$\Delta T = -(\dfrac M 2 + \dfrac {5m}{3})Lg$$ $$\dfrac 1 2 ((M/3 + m)L^2)(\omega^2_f-\omega^2_0)=-(\dfrac M 2 + \dfrac {5m}{3})Lg$$ Finally, we solve for ## \omega_f ##, which is the angular velocity of the vertical rod. $$\omega_f^2=\dfrac {-( M + \dfrac {10m}{3})}{ (M/3 + m)} \dfrac g L+\omega_0^2$$ The negative sign indicates that the angular velocity should decrease, which is logical.
 
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  • #13
The problem has not given you ##\omega_0##, yet it asks you to find ##\omega## when the rod is vertical. In the absence of the initial conditions, most people (myself included) would assume the simplest case, namely that the rod is released from rest at the horizontal 3 o'clock position and swings clockwise to the vertical 6 o'clock position.

On edit: I just noticed that the title states that the rod rotates down but that is not in the statement of the problem.
 
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  • #14
The title of the question says “... A rod with two masses rotates down from a horizontal starting position”.

So the initial conditions are ##\omega_0 = 0## and the rod is horizontal.

This information has unfortunately been omitted from the problem-statement.
 
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  • #15
kuruman said:
The problem has not given you ##\omega_0##, yet it asks you to find ##\omega## when the rod is vertical. In the absence of the initial conditions, most people (myself included) would assume the simplest case, namely that the rod is released from rest at the horizontal 3 o'clock position and swings clockwise to the vertical 6 o'clock position.

On edit: I just noticed that the title states that the rod rotates down but that is not in the statement of the problem.
… and part b specifies "when the rod is vertical after rotating 90 degrees".
MatinSAR said:
Didn't you forget ##\sin \theta##?
As I read the question, part a asks for the angular acceleration when released from horizontal and part b asks for the angular velocity at the vertical. Neither requires you to consider an arbitrary angle in between.
 
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  • #16
kuruman said:
On edit: I just noticed that the title states that the rod rotates down but that is not in the statement of the problem.
The title is edited by a mentor. But I think it is logical that the rod rotates down. What's the problem?!
Steve4Physics said:
The title of the question says “... A rod with two masses rotates down from a horizontal starting position”.

So the initial conditions are ##\omega_0 = 0## and the rod is horizontal.

This information has unfortunately been omitted from the problem-statement.
Actually this information doesn't exist in the problem-statement. The title is edited by a mentor.
haruspex said:
… and part b specifies "when the rod is vertical after rotating 90 degrees".
I think it can rotate downward, Am I wrong? Then my answer at post #12 should be wrong.
haruspex said:
As I read the question, part a asks for the angular acceleration when released from horizontal and part b asks for the angular velocity at the vertical. Neither requires you to consider an arbitrary angle in between.
So the acceleration is changing by time and not constatnt ... Thank you.
 
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  • #17
MatinSAR said:
I think it can rotate downward,
Yes, it starts horizontal (part a) then swings down around the pivot to become vertical (part b), and continues, presumably.
 
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  • #18
haruspex said:
Yes, it starts horizontal (part a) then swings down around the pivot to become vertical (part b), and continues, presumably.
If I want to find the acceleration at an arbitrary point, should I simply include the ##\sin \theta## in the equations?
What do you think about using Euler equations to solve the problem?
 
  • #21
MatinSAR said:
In part a, nothing is rotating yet, so Euler's equation reduces to ##\vec\tau=I\dot{\vec\omega}##.
Part b is solved most simply using energy.
If you mean the generalisation of being at angle ##\theta##, that should work, but it still seems like overkill.
 
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  • #22
haruspex said:
In part a, nothing is rotating yet, so Euler's equation reduces to ##\vec\tau=I\dot{\vec\omega}##.
Part b is solved most simply using energy.
If you mean the generalisation of being at angle ##\theta##, that should work, but it still seems like overkill.
Thank you for your explanation, @haruspex. I understand now. I've asked the TA, and the question specifically wants us to determine the acceleration when the rod is in a horizontal position, as you said.

@kuruman and @Steve4Physics Thank you for your time.
 
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