Inclined Plane Rotational Motion

[phoenixXL]

Homework Statement

A cyclinder of mass m and radius r is rotated about its axis by an angular velocity ω and then lowered gently on an inclined plane as shown in figure. Then:

(a) It will start going upward.
(b) It will first go up and then downward.
(c) It will go downward just after it is lowered.
(d) It can never go upward.

2. The attempt at a solution

Taking torque that is due to sine component of weight about the point of contact,
$$\tau\ =\ I\alpha\\ \implies \tau\ =\ \frac{3}{2}mr^2(\alpha)\\ \implies r.mgsin(30°)\ =\ \frac{3}{2}mr^2(\alpha)\\ \implies \alpha\ =\ \frac{g}{3r}~~...(i)$$
The sense of rotation of angular acceleration and of angular velocity is opposite, therefore soon the cyclinder will stop rotating.

More Equations, from Newtons laws of motion
$$mgsin(30°)\ -\ f\ =\ ma~~...(ii),~~~~f\ =\ frictional force,~~a\ =\ net acceleration.$$
Taking torque about center of mass
$$\tau\ =\ I\alpha\\ \implies f\ =\ \frac{1}{2}mr^2.\frac{g}{3r}~~~~from (i)\\ \implies f\ =\ \frac{mg}{6}~~...(iii)$$

Substituting (iii) in (ii), We get
$$a\ =\ \frac{g}{3}$$

Therefore, we can say that the body when lowered will begin to move down with its rotational motion getting reduced. Therefore, the answer should be (c)

But,
Solution Given in Book

(d) It can never go upward.

Reason: Since net force along the incline is zero, so cyclinder will remain in position till it stops rotating. After that it will start moving downwards.Please help me out, I am not able to clear my concepts of rotational motion.
Thanks for your time

 

[voko]

You do not need torques to solve this problem. All you need to determine is the motion of the centre of mass. The motion of the centre of mass is determined by taking into account all the external forces that act on the system.

What are they?

 

[phoenixXL]

The forces that act on the cyclinder are
- Normal due to Incline
- Weight of the cyclinder
- Friction by virtue of its rotation

But how do we combine them and get it where the body will move thereafter?

It can be seen that
$$\mu\ >\ \frac{c}{1+c}tan30°$$
Therefore it is a case of pure rolling, hence the velocity of the cyclinder would then be
$$v\ =\ \frac{\omega}{r}$$
Therefore it should go up, am I correct ?

 

[voko]

Yes, those are correct forces. I recommend that you write Newton's second law in the components parallel and perpendicular to the incline.

 

[phoenixXL]

Parallel to incline
$$mgsin(30°)\ -\ f_s\ =\ ma$$
Perpendicular to incline
$$N\ =\ mgcos30°$$

Then with friction
$$f_s\ <=\ μN\\ \implies\ f_s\ <=\ μmgcos30°$$

 

[voko]

And what acceleration does that give you?

Is it possible that $$ \mu = \sqrt {\frac 1 3} $$?

 

[phoenixXL]

Yes,
$$\mu\ =\ \frac{1}{\sqrt{3}}$$

It gives zero acceleration.
But what does this infer ?

 

[voko]

What is the initial velocity of the cylinder's centre of mass when it is lowered onto the incline?

 

[phoenixXL]

There is no initial velocity of the centre of mass initially, it only has initial angular velocity.

 

[voko]

So, if the centre of mass is initially stationary, and then the acceleration is zero, does it move anywhere?

 

[phoenixXL]

But only one thing that pinchs, we haven't used the condition of the angular velocity ω.
The conditions we took would be the same as if the one when someone would simply keep a non-rotating cyclinder over the incline,

Solution Given in Book

(d) It can never go upward.

Reason: Since net force along the incline is zero, so cyclinder will remain in position till it stops rotating. After that it will start moving downwards.

Additionally, the book says that it would down once it stops rotating. So, rotation should make some difference, isn't it ?

Thanks for your time

 

[voko]

But only one thing that pinchs, we haven't used the condition of the angular velocity ω.

You did, but implicitly. Because of the rotation, the force of kinetic friction is up the incline.

The conditions we took would be the same as if the one when someone would simply keep a non-rotating cyclinder over the incline

No, that is not true. A non-rotating cylinder would begin to roll down immediately.

 

[phoenixXL]

Right. Thank you so much

 

[Vibhor]

@Voko : I think the cylinder will roll up the incline only if μ >1/√3 i.e say if μ= 0.7 . In that case it will roll up till the angular velocity reduces to zero .It will halt for a a moment and , after that it will roll down .

What is your opinion ?

 

[voko]

It will roll up when the coefficient of kinetic friction satisfies your condition. But it is not correct to say that as soon as its angular velocity is zero it will stop.

 

[Vibhor]

But it is not correct to say that as soon as its angular velocity is zero it will stop.

The friction will gradually decrease as the cylinder rolls up the incline because as the angular speed decreases,there would be less slipping .Eventually friction would become zero when the cylinder stops rotating.After that again friction comes in the scene .This time cylinder rotates clockwise down the incline .

Is it correct ?

 

[phoenixXL]

I missed one point

You did, but implicitly. Because of the rotation, the force of kinetic friction is up the incline.

As it is pure rolling, the contact point must be stationary, hence it must be static friction.

Moreover, if you we just keep a non-rotating cyclinder then also the friction of friction is up the incline as the body tends to move down due to its weight. Am I correct ?

 

[voko]

The friction will gradually decrease

No, this is not correct. Friction does not behave like that.

 

[Vibhor]

No, this is not correct. Friction does not behave like that.

Fine .Thank you.

Then why would the cylinder move up the incline even when it stops rotating .Is it because of the momentum gained ?

When it stops rotating isn't the net force mgsinθ down the incline ? Please explain the motion of the cylinder under the condition μ >1/√3 .

 

[voko]

I missed one point
As the it is pure rolling, the contact point must be stationary, hence it must be static friction.

How is it pure rolling? When you put the cylinder on the incline, the point of contact clearly has some non-zero velocity.

Moreover, if you we just keep a non-rotating cyclinder then also the friction of friction is up the incline as the body tends to move down due to its weight. Am I correct ?

Generally, correct. Details are complicated, however.

 

[voko]

Fine .Thank you.

Then why would the cylinder move up the incline even when it stops rotating .Is it because of the momentum gained ?

When it stops rotating isn't the net force mgsinθ down the incline ? Please explain the motion of the cylinder under the condition μ >1/√3 .

I have thought about this more, and I now think you original assessment was correct. What happens here is that while the cylinder is slipping, it will be accelerating up the incline, while its rotational motion will be slowing down. At some point that will cause it to stop slipping, at which point it will begin to accelerate down, while still rolling up for a while. Then it will come to a complete stop and then will roll down.

My mistake was in not realising that slipping will have to stop before angular velocity becomes zero.

 

[phoenixXL]

How is it pure rolling? When you put the cylinder on the incline, the point of contact clearly has some non-zero velocity.

If the point has some velocity then it should move, as net acceleration is zero.

 

[voko]

If the point has some velocity then it should move, as net acceleration is zero.

It does move - in a circle. The net acceleration on that point is not zero.

 

[phoenixXL]

Ok let me sum up.

- The net acceleration of the center of mass is zero, therefore body doesn't move.
- Every point on the cyclinder has velocity by virtue of its rotational motion, except the center of mass as its not moving.
- The acceleration on the point of contact is due to friction as the point of contact has relative motion with the surface in contact.

This solves the problem. I just hope its correct.

 

[voko]

- The acceleration on the point of contact is due to friction as the point of contact has relative motion with the surface in contact.

This one is not correct. The acceleration on the point of contact is not only due to friction. Imagine a cylinder suspended on a bearing and not touching anything except the bearing itself, about which it can rotate freely. When it rotate, all points of the cylinder that are not on the axis of rotate experience acceleration.

 

[phoenixXL]

Oh, right, the acceleration is also caused due to the centripetal force(rotation). Acceleration because of change of direction of velocity.

So the acceleration of all the points is due to centripetal force, and the acceleration of the point of contact is due to centripetal as well as frictional force.

 

[Tanya Sharma]

Oh, right, the acceleration is also caused due to the centripetal force(rotation). Acceleration because of change of direction of velocity.

So the acceleration of all the points is due to centripetal force, and the acceleration of the point of contact is due to centripetal as well as frictional force.

Hello phoenixXL

Let P be any point on the rotating body with C as the contact point .

$\vec{a}_{P} = \vec{a}_{P,COM} + \vec{a}_{COM}$

Acceleration of COM is given by

$\vec{a}_{COM} = \frac{\vec{F}_{net}}{M}$

Acceleration of the point P with respect to COM is

$\vec{a}_{P,COM} = \vec{a}_{tangential} + \vec{a}_{radial}$

where $\vec{a}_{tangential} = \alpha R$

and $\vec{a}_{radial} = \frac{v^2}{R}$

So, $\vec{a}_{P} = \vec{a}_{tangential} + \vec{a}_{radial} + \vec{a}_{COM}$

If the body is rolling without slipping ,then for the contact point C , $\vec{a}_{tangential} = -\vec{a}_{COM}$

So the total acceleration of point C is given by $\vec{a}_{C} = \vec{a}_{radial}$

Hope that helps

 

[phoenixXL]

Hello phoenixXL

Let P be any point on the rotating body with C as the contact point .

$\vec{a}_{P} = \vec{a}_{P,COM} + \vec{a}_{COM}$

Acceleration of COM is given by

$\vec{a}_{COM} = \frac{\vec{F}_{net}}{M}$

Acceleration of the point P with respect to COM is

$\vec{a}_{P,COM} = \vec{a}_{tangential} + \vec{a}_{radial}$

where $\vec{a}_{tangential} = \alpha R$

and $\vec{a}_{radial} = \frac{v^2}{R}$

So, $\vec{a}_{P} = \vec{a}_{tangential} + \vec{a}_{radial} + \vec{a}_{COM}$

If the body is rolling without slipping ,then for the contact point C , $\vec{a}_{tangential} = -\vec{a}_{COM}$

So the total acceleration of point C is given by $\vec{a}_{C} = \vec{a}_{radial}$

Hope that helps

Thank you Tanya, its crystal clear and lucid now. Thanks a lot