A cylinder connected to a hanged mass

[Robeurer]

Homework Statement

A system consists of a cylinder and a hanging mass. On the left side of the system is the cylinder that is in contact with a wall to its right and floor. The cylinder is also connected to a rope that connects to a pulley and a mass. Both the pulley and the mass are on the right side. The cylinder has a mass of 10 kg and a radius of 4 m while the hanging mass has a mass of 10 kg. Both the wall and the floor have the same coefficient of kinetic friction of 0.5. The mass fell. While the mass fell down, the cylinder is still in contact with both the wall and the floor. What is the system's acceleration? (g = 10m/s^2, the rope and the pulley mass can be neglected, also with the pulley's mass. The cylinder is homogenous).

Relevant Equations

F = ma, torque

Let's declare that for the cylinder,
mass = M = 10 kg
Radius = R = 4 m

For the wall and the floor,
Friction coeff = $\mu$ = 0.5

For the hanging mass,
mass = m = 11 kg

First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on.

• Force on the hanging mass
  $$mg - T = ma$$
• Force(Cylinder) on y
  $$N_f + f_w - Mg = 0$$
• Force(Cylinder) on x
  $$T + f_f - N_w = Ma$$

There's also a torque acting on the cylinder
$$TR - f_w - f_f = I{\alpha}$$
Symplified
$$T - f_w - f_f = \frac{1}{2}Ma$$

That's all what I can say for now. The part that make me confused the most is how to decide the normal force and the friction. What I know that $f <= \mu N$.

That's all

 

[wrobel]

You are said that the system moves so the friction force$=\mu N$

"system's acceleration" is an unclear phrase by the way. A point can have acceleration.

 

[Robeurer]

What I mean by system's acceleration is the acceleration for both the falling mass and for the cylinder.
I stated that the force on the hanging mass is $mg - T = ma$ and $T+f_f-N_w=ma$. The "a" on the right side of the equations is the one that we are looking about.

 

[A.T.]

The part that make me confused the most is how to decide the normal force

The center of mass of the cylinder is not accelerating, so all the forces on the cylinder must cancel.

and the friction. What I know that $f <= \mu N$.

That would be static friction, but you are told the system is not static.

 

[Steve4Physics]

@Robeurer, in addition to what's already been said:

Is the rope wound around the cylinder (like a yo-yo) or is the left end of the rope fixed (to the top of the cylinder)? Almost certainly a yo-yo is intended but this needs to be clear.

What I mean by system's acceleration is the acceleration for both the falling mass and for the cylinder.
I stated that the force on the hanging mass is $mg - T = ma$ and $T+f_f-N_w=ma$. The "a" on the right side of the equations is the one that we are looking about.

Have you considered the possibility that the cylinder and falling mass have different accelerations? I'd guess that you want the acceleration of the hanging mass, not that of the system.

For the hanging mass,
mass = m = 11 kg

Is that a mistake? The hanging mass seems to have changed from 10 kg to 11 kg.

• Force(Cylinder) on y
  $$N_f + f_w - Mg = 0$$

Presumably $N_f$ is the normal force of the floor on the cylinder. And $f_w$ is the frictional force of the wall on the cylinder. A free-body of the cylinder with all forces clearly labelled is recommended.

Since the right hand side of above equation is zero, you are saying that the centre of mass of the cylinder does not accelerate vertically. Is that your intention?

• Force(Cylinder) on x
  $$T + f_f - N_w = Ma$$

The right hand side is non-zero so you are considering the possibility that the cylinder can accelerate in the x-direction. Since the wall prevents the cylinder from moving right, do you mean that you think the cylinder's centre of mass could accelerate left?

There's also a torque acting on the cylinder
$$TR - f_w - f_f = I{\alpha}$$

$f_w$ and $f_f$ also need mutiplying by $R$.

Symplified
$$T - f_w - f_f = \frac{1}{2}Ma$$

You mean 'simplified'. No idea what's happening here!

Minor edit.

 

[ibrkic0]

I stumbled on this thread and thought I'd try to work through it since it looks like you're still figuring out the acceleration. -cool setup with the cylinder against the wall and floor ., reminds me of those Atwood machine variations but with rolling and friction thrown in. I'll assume kinetic friction since the mass is falling, so f = μN for both surfaces, and the cylinder rolls without slipping (from the torque eq9.


given: M=10kg (cylinder),
m=11kg (hanging mass),
μ=0.5, g=10m/s²,
R=4m (but R cancels out anyway)

Equations (based on yours, with directions: f_w upward on wall, f_f to the left on floor helping T pull right):

1. For hanging mass: T = m(g - a) = 110 - 11a
2. Cylinder vertical: N_f = Mg - f_w = 100 - f_w
3. Cylinder horizontal: T + f_f - N_w = Ma = 10a
4. Torque: T - f_w - f_f = (1/2)Ma = 5a (assuming clockwise rotation or whatever matches)

Now, [Too much explicit work has been redacted by the Mentors]

hmm,seems consistent, positive values. If the frictions were in other directions, it might not work, but this seems to fit with the cylinder rolling rightward as the mass falls.

Does this make sense? Or maybe I have the mass wrong ...you said m=10kg initially but equations have mg=110, so probably 11kg. Let me know if the signs are off or if there's a diagram.

 

[Robeurer]

@Steve4Physics @ibrkic0 Looks like there's a typo. The mass of the hanging object is 11 kg

 

[Steve4Physics]

I stumbled on this thread and thought I'd try to work through it since it looks like you're still figuring out the acceleration.

Hi @ibrkic0. Welcome to PF.

This forum tries to provide guidance/help with homework problems - but not complete solutions. The aim is to guide/support the OP so they can solve the problem for themself. E.g. see Item 8. in Homework Help Guidelines for Students and Helpers.

Cylinder horizontal: T + f_f - N_w = Ma = 10a

The acceleration (a) in this equation is the (x-component of) the acceleration of the cylinder’s centre of mass. Of course, if the cylinder only rotates about its axis, then this acceleration value is zero.

Torque: T - f_w - f_f = (1/2)Ma = 5a (assuming clockwise rotation or whatever matches)

If I'm interpreting the above equation correctly, the acceleration (a) in the equation is the acceleration** of a point on the edge of the cylinder. So the symbol 'a' is being used for two completely different accelerations. This is the same mistake as the OP makes in Post #1.

**Edit. "acceleration" should be replaced by "tangential component of acceleration", as there is also a centripetal (radial) component. Thanks to @wrobel for pointing this out.

 

[wrobel]

(a) in the equation is the acceleration of a point on the edge of the cylinder.

the tangent component of the acceleration of a point on the edge

 

[Robeurer]

What is the system's acceleration?

@Steve4Physics @ibrkic0 I'm really really sorry that it looks like there's a mistake. The question asks about the acceleration of the falling mass and does not mention anything about the system's acceleration. Sorry.

 

[Robeurer]

The center of mass of the cylinder is not accelerating, so all the forces on the cylinder must cancel.

How come the cylinder is not accelerating? I thought that it will also accelerate just like the falling mass like in the Atwood's machine.

If the cylinder is not accelerating, all we need to do to find the acceleration of the falling mass is by finding the value of $T$. Is it correct?

 

[Robeurer]

Is the rope wound around the cylinder (like a yo-yo) or is the left end of the rope fixed (to the top of the cylinder)? Almost certainly a yo-yo is intended but this needs to be clear.

I'm not sure either, but maybe it's wound around the cylinder.

 

[Robeurer]

I'm not sure either, but maybe it's wound around the cylinder.

After re-read the question, the rope is wound around it.

 

[haruspex]

How come the cylinder is not accelerating?

The horizontal forces on it are the friction from the floor, the pull of the rope, and the normal force from the wall.
The first two point to the right. A normal force between two rigid bodies in contact is the minimum force needed to prevent interpenetration. Since the wall is stationary, it will be just sufficient to balance the first two.
The same logic applies to the force from the floor.

 

[A.T.]

How come the cylinder is not accelerating?

Read what I wrote exactly:

The center of mass of the cylinder is not accelerating,