# How does the system move?

Nexus99
Homework Statement:
A solid cylinder with mass M and radius R is held suspended in a vertical plane thanks to a pin P passing for its axis.
The cylinder is free to move with negligible friction around the pin. On the side of the cylinder
a vertical panel is pushed horizontally against the cylinder by an ideal spring which provides a
elastic force ##\vec{F_{el}} ##. A pulley of negligible mass and radius r is applied to the cylinder axis and moves with the cylinder. The pulley is connected through an ideal wire to a mass m. Between the vertical panel and the cylinder there is friction with coefficients ## \mu_s ## (static coefficient) and ## \mu_k ## (kinetic coefficient)
Relevant Equations:
nothing Can't visualize how the system moves, can anyone help me?

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Homework Statement:: A solid cylinder with mass M and radius R is held suspended in a vertical plane thanks to a pin P passing for its axis.
The cylinder is free to move with negligible friction around the pin. On the side of the cylinder
a vertical panel is pushed horizontally against the cylinder by an ideal spring which provides a
elastic force ##\vec{F_{el}} ##. A pulley of negligible mass and radius r is applied to the cylinder axis and moves with the cylinder. The pulley is connected through an ideal wire to a mass m. Between the vertical panel and the cylinder there is friction with coefficients ## \mu_s ## (static coefficient) and ## \mu_k ## (kinetic coefficient)
Relevant Equations:: nothing

View attachment 267605
Can't visualize how the system moves, can anyone help me?
Perhaps it doesn’t. What equilibrium equation could you write?

I don't have the physical insight of @haruspex, so I wondered if you could clarify a few things... can the pin ##P## translate horizontally? Also where is the other end of the string attached to - is it just fixed to some point on the pulley, and the string is wound around a few times?

The equilibrium case looks like just a matter of taking torques about the centre of the disk, and resolving forces on the disk (not forgetting the contact force from the pin!).

Nexus99
I don't have the physical insight of @haruspex, so I wondered if you could clarify a few things... can the pin ##P## translate horizontally?
I don't think it can

Also where is the other end of the string attached to - is it just fixed to some point on the pulley, and the string is wound around a few times?
Yes

Perhaps it doesn’t. What equilibrium equation could you write?
Traslation equilibrium:
mass m: ##T - mg = 0##
mass M:
x : ## F_{el} - N_x = 0##
y: ## N_y - T - Mg - F_k = 0 ##

Rotational equilibrium:
## Tr - F_k R = 0 ##

So long as your ##\hat{x}## basis vector points to the left, your equations are good.

Maybe now you can tweak your equations for the dynamic case?

Nexus99
So long as your ##\hat{x}## basis vector points to the left, your equations are good.

Maybe now you can tweak your equations for the dynamic case?
My ##\hat{x}## vector points to te right.

I thought that the elastic force pushed the cylinder to the right and the constraint reaction force should oppose it, pushing to the left. Isn't true?

My ##\hat{x}## vector points to te right.

I thought that the elastic force pushed the cylinder to the right and the constraint reaction force should oppose it, pushing to the left. Isn't true?

Okay, but if your ##\hat{x}## vector points to the right then ##N_x## will be positive if ##\vec{N}## points to the right and negative if ##\vec{N}## points to the left, i.e. ##\vec{N} = N_x \hat{x}##. That is to say that your equation would need to be$$F_{el} + N_x = 0$$Alternatively you could write ##N_x = -N##, so that$$F_{el} - N = 0$$

Nexus99
Okay, but if your ##\hat{x}## vector points to the right then ##N_x## will be positive if ##\vec{N}## points to the right and negative if ##\vec{N}## points to the left, i.e. ##\vec{N} = N_x \hat{x}##. That is to say that your equation would need to be$$F_{el} + N_x = 0$$Alternatively you could write ##N_x = -N##, so that$$F_{el} - N = 0$$
Ok i got it

Anyway the equations for dynamic case are:
Traslation:
mass m: ##T - mg = -ma##
mass M:
x : ## F_{el} + N_x = 0##
y: ## N_y - T - Mg - F_k = 0 ##

Rotational:
## Tr - F_k R = I_P \alpha = I_p \frac{a}{r} ## (not sure if it is ##\frac{a}{r}## or ##\frac{a}{R} ##)

• etotheipi
Anyway the equations for dynamic case are:
Traslation:
mass m: ##T - mg = -ma##
mass M:
x : ## F_{el} + N_x = 0##
y: ## N_y - T - Mg - F_k = 0 ##

Rotational:
## Tr - F_k R = I_P \alpha = I_p \frac{a}{r} ## (not sure if it is ##\frac{a}{r}## or ##\frac{a}{R} ##)

Yes I think that's perfect. And you're right that it must be ##\frac{a}{r}##, since you require that the ##y## components of the acceleration of the point on the pulley and suspended mass above and below each other are equal, i.e.$$\vec{\alpha} \times r\hat{x} = -a\hat{y}$$ $$-\alpha r \hat{y} = -a\hat{y} \implies \alpha = \frac{a}{r}$$

• Nexus99
Nexus99
Yes I think that's perfect. And you're right that it must be ##\frac{a}{r}##, since you require that the ##y## components of the acceleration of the point on the pulley and suspended mass above and below each other are equal, i.e.$$\vec{\alpha} \times r\hat{x} = -a\hat{y}$$ $$-\alpha r \hat{y} = -a\hat{y} \implies \alpha = \frac{a}{r}$$
Ok thanks,
I have a doubt now, is ## N_x ## only the force exerted by the pin or it contains also the force exerted by the panel?
And i still can't visualize clearly how the system is moving.
I think that the gravitational force push the mass m to the down, so the cylinder rotates clockwise. But, after that the wire has unwound completely, how can it rewind around the pulley?

##\vec{N}## is the force exerted by the pin on the cylinder. The force exerted by the panel on the cylinder is ##\vec{F}_{el}## (you can see this from Newton's third law; the spring exerts ##\vec{F}_{el}## on the panel, so for order for the panel to be in equilibrium the cylinder must exert ##-\vec{F}_{el}## on the panel. It follows from NIII that the panel exerts ##\vec{F}_{el}## on the cylinder).

As far as I can tell, the mass will accelerate downward and the cylinder will begin accelerating rotationally. I think the problem only wants you to consider the interval in which the wire is still unravelling, because when the wire reaches maximum length the point of attachment of the string will start rotating with the cylinder and the suspended mass will start to be pulled horizontally in some chaotic fashion.

Nexus99
##\vec{N}## is the force exerted by the pin on the cylinder. The force exerted by the panel on the cylinder is ##\vec{F}_{el}## (you can see this from Newton's third law; the spring exerts ##\vec{F}_{el}## on the panel, so for order for the panel to be in equilibrium the cylinder must exert ##-\vec{F}_{el}## on the panel. It follows from NIII that the panel exerts ##\vec{F}_{el}## on the cylinder).

As far as I can tell, the mass will accelerate downward and the cylinder will begin accelerating rotationally. I think the problem only wants you to consider the interval in which the wire is still unravelling, because when the wire reaches maximum length the point of attachment of the string will start rotating with the cylinder and the suspended mass will start to be pulled horizontally in some chaotic fashion.
You're right, thanks for help

I have some doubts on this question:
The work of the friction force after the mass m has dropped by a height d

L = ## \int_{0}^{\theta'} \tau_{F{k}} \, d \theta ##
i know that:
## \frac{d}{r} = \theta' ##
and
## \tau_{F{k}} = - R \mu_k F_{el} ##
So:
##L = -\frac{d}{r} R \mu_k F_{el} ##
is it right?

Last edited:
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You're right, thanks for help

I have some doubts on this question:
The work of the friction force after the mass m has dropped by a height d

L = ## \int_{0}^{\theta'} \tau_{F{k}} \, d \theta ##
i know that:
## \frac{d}{r} = \theta' ##
and
## \tau_{F{k}} = - R \mu_k F_{el} ##
So:
##L = -\frac{d}{r} R \mu_k F_{el} ##
is it right?
Yes, but you can there more directly: force = ##\mu_k F_{el} ##, relative distance moved by the surfaces in contact = ##d\frac{R}{r} ##

• Nexus99