Stick leaning on the wall, find the acceleration from the initial position

mattlfang
Homework Statement:
Stick with mass m leans on the wall on B end. A end hinges on the center of the cylinder with mass M and radius r. Initial position has angle of 45 degrees. No friction on the wall or the hinges. Cylinder starts rolling from initial position, but the frictional force on the cylinder can be ignored. What's the acceleration of A point at the initial position
Relevant Equations:
Decomposition of forces. Newtonian second law.
So the acceleration of point A was given by a force F exerted on cylinder that's along the direction of the stick, decomposed into the horizontal direction. so aA = F cos Θ

The same force along the opposite direction is exerted on stick, and if we decompose that in vertical and horizontal direct, we can compute the acceleration of the stick m. so aStick = (F cos Θ - mg) / m

The acceleration of the stick and the cylinder satisfies certain geometric relation, then we can probably compute the force and the acceleration.

But I struggle to finish the last part off.

this is the answer. but I have no solutions.

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Mentor
You may want to draw a Free Body Diagram for the stick and for the cylinder.

mattlfang
does the problem give a mass and radius for the cylinder
As the statement suggests, "cylinder with mass M and radius r." But I actually don't think the radius is relevant.

Gold Member
No friction anywhere. Cylinder starts rolling from initial position.
Mutually contradictory statements. I presume there is static friction between the cylinder and the surface upon which it rolls, else it would just start sliding, not rolling.

You have some text, but have not expressed the component forces acting on/by the stick at points A and B, and for that matter at point C where the cylinder meets the surface.

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Delta2, SammyS and Chestermiller
Mentor
Mutually contradictory statements. I presume there is static friction between the cylinder and the surface upon which it rolls, else it would just start sliding, not rolling.
Oh, good point! I didn't even think of that.

@mattlfang -- is there friction on the bottom surface to let the cylinder roll, or is there "no friction anywhere" so the cylinder just slides?

mattlfang
Mutually contradictory statements. I presume there is static friction between the cylinder and the surface upon which it rolls, else it would just start sliding, not rolling.

You have some text, but have not expressed the component forces acting on/by the stick at points A and B. That's the starting point, after which the problem reduces to a simpler case.
I have added more equations. Yes. I meant "No friction on the wall or the hinges. Cylinder starts rolling from initial position, but the frictional force on the cylinder can be ignored"

Mentor
but the frictional force on the cylinder can be ignored"
But not its Moment of Inertia...

Gold Member
Admittedly not a trivial problem. The horizontal force exerted on the cylinder by the stick is not equal to the force exerted on the stick by the wall. The stick gains both momentum (west) and angular momentum (clockwise) as it falls, both reducing the acceleration of the cylinder from the much simpler case of a massless stick with a spring providing its 'weight' at the midpoint.
So the equations involved need to include all of these components. At least we only have to compute initial acceleration, so we can leave the calculus at home.

berkeman
mattlfang
Admittedly not a trivial problem. The horizontal force exerted on the cylinder by the stick is not equal to the force exerted on the stick by the wall. The stick gains both momentum (west) and angular momentum (clockwise) as it falls, both reducing the acceleration of the cylinder from the much simpler case of a massless stick with a spring providing its 'weight' at the midpoint.
So the equations involved need to include all of these components. At least we only have to compute initial acceleration, so we can leave the calculus at home.

Can you elaborate on how to model the angular momentum here?

Gold Member
A (presumed solid) cylinder with a known mass and radius has an expressible moment of inertia.
Likewise a (presumed uniform density) stick with known length and mass.
Eww, that's a problem since length of stick is not provided. Is the answer not dependent on stick length then? A long stick will have more moment but less angular acceleration, so it might cancel out. I haven't actually worked out the answer.

I would work backwards, starting with presuming a known acceleration (to west, of point A), and from that computing horizontal force at A. Now given that force, compute mass of stick with known length 'd'. If the mass m turns out to be independent of length d, then all you have to do is solve for acceleration given the computed m. If it is a function of d, then the problem cannot be solved without that being known.

mattlfang
A (presumed solid) cylinder with a known mass and radius has an expressible moment of inertia.
Likewise a (presumed uniform density) stick with known length and mass.
Eww, that's a problem since length of stick is not provided. Is the answer not dependent on stick length then? A long stick will have more moment but less angular acceleration, so it might cancel out. I haven't actually worked out the answer.

I would work backwards, starting with presuming a known acceleration (to west, of point A), and from that computing horizontal force at A. Now given that force, compute mass of stick with known length 'd'. If the mass m turns out to be independent of length d, then all you have to do is solve for acceleration given the computed m. If it is a function of d, then the problem cannot be solved without that being known.
Thanks. I posted the answer, but I don't have the solutions. Surprisingly simple actually.

Can we actually argue the angular momentum conserves here? The whole system is being exerted the gravity force, no?

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Gold Member
Can we actually argue the angular momentum conserves here? The whole system is being exerted the gravity force, no?
No, there is no conservation here. The system isn't closed since unbalanced torque and linear forces are supplied by the wall (at B) and the ground where the rolling takes place.

ergospherical
Hm, I tried conserving the energy ##E = E(\theta, \dot{\theta})## of the system in order to derive the equations of motion, but then arrived at a result of ##a_A = \dfrac{mg}{3M+m}## [edit: this result neglects rotational KE of rod, see #16]. Either I made a numerical slip, or the author used different assumptions!

It's indeed strange that the author specifies no friction, when in fact friction at the contact point between the cylinder and the ground is what must cause the cylinder to go into rotation. It would, however, do no work.

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mattlfang
Hm, I tried conserving the energy ##E = E(\theta, \dot{\theta})## of the system in order to derive the equations of motion, but then arrived at a result of ##a_A = \dfrac{mg}{4M+m}##. Either I made a numerical slip, or the author used different assumptions!

It's indeed strange that the author specifies no friction, when in fact friction at the contact point between the cylinder and the ground is what must cause the cylinder to go into rotation. It would, however, do no work.

Do you mind show a bit of your work?

ergospherical
Apologies, I neglected the rotational KE of the rod in my previous post (now removed anyway due to PF guidelines).

berkeman
ergospherical
I'll re-post the first part of the post (now with rotational KE of rod included ) to set you on the right track, and I'll leave it to you to finish it off!

With the origin in the corner and denoting the elevation angle ##\theta##, then the centre of the rod has position and velocity:\begin{align*}

x_1 = \dfrac{1}{2}l\cos{\theta} &\implies \dot{x}_1 = - \dfrac{1}{2}l \dot{\theta} \sin{\theta} \\

y_1 = r + \dfrac{1}{2}l \sin{\theta} &\implies \dot{y}_1 = \dfrac{1}{2}l \dot{\theta} \cos{\theta}

\end{align*}so the speed of the centre of the rod satisfies ##v^2 = \dfrac{1}{4}l^2 \dot{\theta}^2##. Meanwhile the position of the centre of the cylinder is ##x_2 = l\cos{\theta}##, so its speed satisfies ##v_2^2 = l^2 \dot{\theta}^2 \sin^2{\theta}##. The rotational KE of the rod is ##\dfrac{1}{24}ml^2 \dot{\theta}^2##. Finally, the cylinder is rolling, so its angular speed satisfies ##\omega^2 = {v_2}^2/r^2 = \dfrac{l^2}{r^2} \dot{\theta}^2 \sin^2{\theta}##. Taking the moment of inertia as ##\dfrac{1}{2}Mr^2##, the rotational KE of the cylinder is ##\dfrac{1}{4}Mr^2 \omega^2##. The total energy is\begin{align*}

E = \dfrac{1}{2}l^2 \dot{\theta}^2(\dfrac{3}{2}M\sin^2{\theta} + \dfrac{1}{3} m ) + \dfrac{1}{2}mgl\sin{\theta}

\end{align*}

Does that all make sense? What's left for you to do is find the time derivative ##\dot{E}## of the energy, equate it to zero, and evaluate all the quantities there at ##t=0##. And finally relate ##\ddot{\theta}## to ##\ddot{x}_2## using a previous equation. Can you finish it off?

Homework Helper
Gold Member
2022 Award
I'll re-post the first part of the post (now with rotational KE of rod included ) to set you on the right track, and I'll leave it to you to finish it off!

Does that all make sense? What's left for you to do is find the time derivative ##\dot{E}## of the energy, equate it to zero, and evaluate all the quantities there at ##t=0##. And finally relate ##\ddot{\theta}## to ##\ddot{x}_2## using a previous equation. Can you finish it off?
The approach using forces, torques and accelerations is not trivial, but it's a bit simpler than the energy conservation algebra above.
Do you mind show a bit of your work?
We haven’t seen any of yours since post #1, where you wrongly assumed the force the stick exerts on the cylinder is along the line of the stick.
Allow for that reaction force to be in any direction.
Define some unknowns and write some equations.

berkeman
Mentor
We haven’t seen any of yours since post #1
Exactly. @mattlfang -- You have been given excellent hints in this thread. You need to start showing some effort or the thread will be closed. It's kind of obvious (to me at least) that you are trying to get others to do as much of your work on this as possible.

ergospherical
mattlfang
Ok, sorry for the delay. Thanks for the above hint on computing the kinetic energy. Let me show more of my work. I managed to get the answer. However I am keen to see the solution that uses "forces, torques and accelerations".

Let O be the point shown below, then at time 0 AB rotates around O. Let C be the midpoint of AB

$$\omega_{AB} = \frac{V_A}{OA} = \frac{v_A}{d \sin \theta}$$
$$v_C = \omega_{AB} OC = \frac{v_A}{2 \sin \theta}$$

At any moment, the kinetic energy

$$E = \frac{1}{2} M v_A^2 + \frac{1}{2} \frac{1}{2} M R^2 ( \frac{v_A}{R} )^2 + \frac{1}{2} m v_C^2 + \frac{1}{2} \frac{1}{12} m d^2 \omega_{AB}^2$$

Power

$$P = - mgv_C \cos \theta$$

Given that ##\frac{dE}{dt} = P##, and ##\frac{d \theta}{ dt } = \omega_{AB}##, and ##\theta = 45## degrees and ##v_A = 0##, we get $$a_A = \frac{3mg}{4m + 9M}$$

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berkeman
Homework Helper
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2022 Award
Ok, sorry for the delay. Thanks for the above hint on computing the kinetic energy. Let me show more of my work. I managed to get the answer. However I am keen to see the solution that uses "forces, torques and accelerations".

Let O be the point shown below, then at time 0 AB rotates around O.

$$\omega_{AB} = \frac{V_A}{OA} = \frac{v_A}{d \sin \theta}$$
$$v_C = \omega_{AB} OC = \frac{v_A}{2 \sin \theta}$$

At any moment, the kinetic energy

$$E = \frac{1}{2} M v_A^2 + \frac{1}{2} \frac{1}{2} M R^2 ( \frac{v_A}{R} )^2 + \frac{1}{2} m v_C^2 + \frac{1}{2} \frac{1}{12} m d^2 \omega_{AB}^2$$

Power

$$P = - mgv_C \cos \theta$$

Given that ##\frac{dE}{dt} = P##, and ##\frac{d \theta}{ dt } = \omega_{AB}##, and ##\theta = 45##degrees and ##v_A = 0##, we get ##a_A = \frac{3mg}{4m + 9M}##
Well done.

With forces, I write
H = horizontal component of force at axle
N = reaction force from wall
##\alpha## = anticlockwise acceleration of cylinder
2L = length of stick
Torque about point of contact with ground:
##\frac 32Mr^2\alpha=Hr##
Acceleration of stick:
##m\frac 12\alpha r=N-H##
So ##N=\frac 12(3M+m)r\alpha##.

Note that the linear acceleration of the mass centre of the stick is towards A (as a fixed point in space), so does not contribute to its angular acceleration about A.
##mgL\cos(\theta)-2NL\sin(\theta)=\frac 13mL^2(\frac{r\alpha \cos(\theta)}L)##
Since sin = cos here:
##mg-2N=\frac 13mr\alpha##
Etc.

Lnewqban
mattlfang
Well done.

With forces, I write
H = horizontal component of force at axle
N = reaction force from wall
##\alpha## = anticlockwise acceleration of cylinder
2L = length of stick
Torque about point of contact with ground:
##\frac 32Mr^2\alpha=Hr##
Acceleration of stick:
##m\frac 12\alpha r=N-H##
So ##N=\frac 12(3M+m)r\alpha##.

Note that the linear acceleration of the mass centre of the stick is towards A (as a fixed point in space), so does not contribute to its angular acceleration about A.
##mgL\cos(\theta)-2NL\sin(\theta)=\frac 13mL^2(\frac{r\alpha \cos(\theta)}L)##
Since sin = cos here:
##mg-2N=\frac 13mr\alpha##
Etc.

Thanks, This is indeed a clever solution that doesn't need calculus.

berkeman and Lnewqban
Homework Helper
Gold Member
Note that the linear acceleration of the mass centre of the stick is towards A (as a fixed point in space), so does not contribute to its angular acceleration about A.
That's a clever observation.
This means that ##\vec \tau_A = \frac {d \vec L_A}{dt} = \frac{d}{dt} \left(m\;\vec r_{_{AC}} \times \vec v_c + I_c \; \vec \omega_s \right)##, where ##\vec \omega_s## is the angular velocity of the stick and ##I_c## is the moment of inertia of the stick about its center of mass. The time derivative of the first term on the right is zero due to your observation. So, we find ##\tau_A = I_c \alpha_s##, where ##\alpha_s## is the angular acceleration of the stick. It might seem odd that the torque equation about ##A## involves ##I_c## rather than ##I_A##, but I think this is right for this situation.

##mgL\cos(\theta)-2NL\sin(\theta)=\frac 13mL^2(\frac{r\alpha \cos(\theta)}L)##
I'm having trouble with the right-hand side of this equation. I get

##I_c \alpha_s = \left(\frac 1 {12} m (2L)^2\right)\left(\frac{r \alpha}{2L\sin\theta}\right) = \left(\frac 1 3 mL^2\right)\left(\frac{r \alpha}{2L\sin\theta}\right)##

So, I get the ##\frac 1 3 m L^2## but I get something else for the last factor which expresses the angular acceleration of the stick in terms of the angular acceleration of the cylinder. If you consider for the moment the case ##\theta = 90^o##, then your expression ##\frac{r\alpha \cos(\theta)}L## would give no angular acceleration of the stick. But I would expect the angular acceleration of the stick in this case to be ##\frac{r \alpha_{\rm cyl}}{2L}##.

Nevertheless, for the case where ##\theta = 45^o##, I do get your final equation ##mg-2N=\frac 13mr\alpha##. So everything still works out.