Question about Irodov 1.258

[adjurovich]

We are asked to find the tension in the rope. First from the first, we can assume that tensions in both ropes are equal, so we can treat them as a single rope since they are wound symmetrically. That “rope” will act tangentially to both cylinders so it exerts torque, the torque equations are:

$Iα_1=RT$
$Iα_2=RT$

We can clearly see that angular accelerations are equal.

By adding these two equations and rearranging them, we get:

$α=\dfrac{4T}{mR}$

It all makes sense. However, I am confused about the accelerations part:

The upper cylinder has only tangential acceleration, and the no-slipping condition says:

$a_{tan} = a_{rope}$

So: $Rα=a_{rope}$

For the lower body, its net acceleration will be:

$a - R \alpha = a_{rope}$

Now, if I got it right, we can treat this acceleration $a$ (translational) as the component of net tangential acceleration since every point has equal acceleration $a$ but so does the point tangential to the cylinder?

And thus by combining these two we get:

$a = 2R\alpha$

I understand the rest so I will stop here. Please correct me if I’m wrong.

 

[A.T.]

View attachment 346996

We are asked to find the tension in the rope. First from the first, we can assume that tensions in both ropes are equal, so we can treat them as a single rope since they are wound symmetrically. That “rope” will act tangentially to both cylinders so it exerts torque, the torque equations are:

$Iα_1=RT$
$Iα_2=RT$

If you combine the two ropes into one, it should be $2T$.

For the lower body, its net acceleration will be:

$a - R \alpha = a_{rope}$

Now, if I got it right, we can treat this acceleration $a$ (translational) as the component of net tangential acceleration since every point has equal acceleration $a$ but so does the point tangential to the cylinder?

If $a = (mg -2T)/m$ then its the acceleration of the center of mass, not of every point on the rotating body.

 

[adjurovich]

If you combine the two ropes into one, it should $2T$.

I forgot to write 2T, I had some trouble with latex! Thanks for remark.

If $a = (mg -2T)/m$ then its the acceleration of the center of mass, not of every point on the rotating body.

Each point of a body that moves translationally has equal velocity (acceleration)? If car moves with 2m/s, each point will travel 2 meters for one second, If I’m correct?

 

[A.T.]

Each point of a body that moves translationally has equal velocity (acceleration)?

The lower body is not just moving translationally, but also rotating. Different points on it have different accelerations.

 

[adjurovich]

The lower body is not just moving translationally, but also rotating. Different points on it have different accelerations.

I understand it, but what boggles my mind is the lower pulley. Acceleration of rope must be equal to the tangential acceleration of upper pulley because of no slipping condition. That must also be the case with lower pulley. However I don’t quite understand what happens with translational acceleration of lower pulley

 

[A.T.]

However I don’t quite understand what happens with translational acceleration of lower pulley

Transnational acceleration of which part of the lower pulley? The vertical acceleration at the center of mass is:
$a = (mg -2T)/m$

But the vertical acceleration at rope lift off point is equal to $a_{rope}$. You have described their relation correctly already:

$a - R \alpha = a_{rope}$

So I'm not sure what is still unclear.

 

[adjurovich]

Transnational acceleration of which part of the lower pulley? The vertical acceleration at the center of mass is:
$a = (mg -2T)/m$

But the vertical acceleration at rope lift off point is equal to $a_{rope}$. You have described their relation correctly already:

So I'm not sure what is still unclear.

It makes sense to me mathematically if put that way, but I wouldn’t be so sure of how to explain it theoretically (physically)

 

[A.T.]

It makes sense to me mathematically if put that way, but I wouldn’t be so sure of how to explain it theoretically (physically)

The acceleration of the center of mass is just Newton'S 2nd Law. The acceleration at rope lift off point follows from the no-slip constraint, and is equivalent to rolling on an accelerating surface.

 

[adjurovich]

The acceleration of the center of mass is just Newton'S 2nd Law. The acceleration at rope lift off point follows from the no-slip constraint, and is equivalent to rolling on an accelerating surface.

What bothers me is this equation:

$a_{rope} = a - a_{tangential}$

How did we exactly come up with this equation? I read the solution but I don’t get it

 

[A.T.]

What bothers me is this equation:

$a_{rope} = a - a_{tangential}$

How did we exactly come up with this equation? I read the solution but I don’t get it

In the center of mass frame, the acceleration of the lifting rope would be $- a_{tangential}$ (we define up as negative). To transform into the the inertial frame you add the acceleration of the frame $a - a_{tangential}$.

 

[adjurovich]

In the center of mass frame, the acceleration of the lifting rope would be $- a_{tangential}$ (we define up as negative). To transform into the the inertial frame you add the acceleration of the frame $a - a_{tangential}$.

This is basically equivalent to relative velocity (relative acceleration)?

 

[berkeman]

Thread closed temporarily for Moderation...

 

[haruspex]

Transnational acceleration

Hurtling down the Khyber pass, for example?

This is basically equivalent to relative velocity (relative acceleration)?

Yes. You can start with a displacement relationship: if the rope descends $\Delta y_{rope}$ and the lower cylinder rotates through $\Delta\theta$ then the cylinder's centre descends $\Delta y_{cyl}=\Delta y_{rope}+R\Delta\theta$ (with a suitable choice of signs).
Differentiate wrt time once to get the velocity relationship, twice for the acceleration.

 

[A.T.]

This is basically equivalent to relative velocity (relative acceleration)?

Yes, a Galilean Transformation is just addition of velocities. But since the derivative of a sum is the sum of derivatives, this also applies to accelerations.

 

[adjurovich]

Yes, a Galilean Transformation is just addition of velocities. But since the derivative of a sum is the sum of derivatives, this also applies to accelerations.

Thanks for help!