Cylinder hitting inclined plane

[Vibhor]

Homework Statement

Homework Equations

The Attempt at a Solution

To bring the cylinder to a stop both linear impulse and angular impulse would be required . Linear impulse would be provided by the normal force from the plane whereas the angular impulse has to be provided by the frictional torque . Assuming the inclined plane has friction , I think both options A) and C) should be correct .

But this is incorrect answer .

If the plane is assumed frictionless then I think it would not be possible to stop the cylinder from rotating irrespective of the angle of the plane .

Any help is very much appreciated .

Thanks

 

[mrnike992]

If the plane is assumed frictionless

Could the plane be assumed frictionless?

 

[TSny]

If the plane is assumed frictionless then I think it would not be possible to stop the cylinder from rotating irrespective of the angle of the plane .

Sounds right.

Can you make any statement about the direction of the net linear impulse felt by the cylinder when it meets the inclined plane?

 

[Vibhor]

Can you make any statement about the direction of the net linear impulse felt by the cylinder when it meets the inclined plane?

If angle of inclined plane is 90° , then linear impulse imparted by plane would be horizontal towards left . If angle of inclined plane is 120° , then linear impulse imparted by plane would make an angle 60° with the vertical .

 

[TSny]

If angle of inclined plane is 90° , then linear impulse imparted by plane would be horizontal towards left . If angle of inclined plane is 120° , then linear impulse imparted by plane would make an angle 60° with the vertical .

Are you referring to the net impulse or just the impulse due to the normal force from the incline? We agreed that friction must act.

In order to suddenly bring the center of mass of the cylinder to rest, what must be the direction of the net impulse?

 

[Vibhor]

Are you speaking about the net impulse or just the impulse due to the normal force from the incline?

Sorry . That was impulse due to normal only .

In order to suddenly bring the center of mass of the cylinder to rest, what must be the direction of the net impulse?

Are you hinting that frictional torque in case of 90° could make the cylinder jump ?

 

[TSny]

Are you hinting that the frictional torque in case of 90° could make the cylinder jump ?

No, I was just concentrating on the direction that the net linear impulse must have in order to stop the linear motion of the cylinder's center of mass.

 

[Vibhor]

No, I was just concentrating on the direction that the net linear impulse must have in order to stop the linear motion of the cylinder's center of mass.

I think we cannot get the direction of the net linear impulse due to friction and normal without knowing their magnitudes .

But don't you think that frictional torque in case of 90 °could indeed make the cylinder move upwards ?

 

[TSny]

I think we cannot get the direction of the net linear impulse due to friction and normal without knowing their magnitudes .

How is the direction of the net impulse related to the direction of the change in linear momentum of the cylinder?

But don't you think that frictional torque in case of 90 °could indeed make the cylinder move upwards ?

Yes, I think that could happen.

 

[Vibhor]

How is the direction of the net impulse related to the direction of the change in linear momentum of the cylinder

They should be in the same direction .

But in case of 90° the net impulse imparted by the plane cannot be in the horizontal direction towards left ( which should be the direction of change of linear momentum of the cylinder ). So, 90° is not possible .

Right ?

 

[TSny]

I think that's right.

 

[Vibhor]

Ok .

But , do you agree that cylinder would definitely lose contact with the floor (move up) if angle is 90° ?

 

[TSny]

I don't think the cylinder would necessarily leave the floor. Got to quit for tonight. I'll check back tomorrow.[emoji42]

 

[haruspex]

Ok .

But , do you agree that cylinder would definitely lose contact with the floor (move up) if angle is 90° ?

if there is friction (and you have proved there must be) then yes.

In order to suddenly bring the center of mass of the cylinder to rest, what must be the direction of the net impulse?

To clarify, that is net of all the impulses involved. Vibhor, what are all the potential impulses?

 

[Vibhor]

Got to quit for tonight. I'll check back tomorrow.

No problem

I don't think the cylinder would necessarily leave the floor.

I think there would be a non zero component of net impulse in vertical direction (due to frictional torque ) which means the center of mass of the cylinder would definitely go up .

Another thing I wanted to clarify is that , I think it is also possible that the cylinder might not stop if the angle is 120° ?? [I know the question is asking whether there is a chance for the cylinder to stop , which it has if angle is 120° , but not if it is 90 °]

 

[haruspex]

I think it is also possible that the cylinder might not stop if the angle is 120° ?? [

It is possible, e.g. if there is insufficient friction. But suppose there is plenty of friction from the sloping surface. What could happen then?

 

[Vibhor]

if there is friction (and you have proved there must be) then yes

Ok . In post#13 TSny is of the opinion that cylinder would not necessarily leave the floor .

 

[Vibhor]

But suppose there is plenty of friction from the sloping surface. What could happen then?

It might roll with/without slipping backwards (towards left ) .

 

[haruspex]

Ok . In post#13 TSny is of the opinion that cylinder would not necessarily leave the floor .

It must. In order to counter the rotational momentum, there must be an upward impulse from the ramp. There is nowhere for a downward impulse to come from, so there will be a net upward impulse.
When the angle exceeds 90 degrees, the normal impulse from the ramp can provide the necessary downward impulse.

 

[Vibhor]

What about post#18 ?

 

[haruspex]

It might roll with/without slipping backwards (towards left ) .

Yes.

 

[TSny]

Ok . In post#13 TSny is of the opinion that cylinder would not necessarily leave the floor .

It must. In order to counter the rotational momentum, there must be an upward impulse from the ramp. There is nowhere for a downward impulse to come from, so there will be a net upward impulse.
When the angle exceeds 90 degrees, the normal impulse from the ramp can provide the necessary downward impulse.

OK. The key phrase is “immediately come to rest”. I see two ways to interpret this

(i) The time interval $\delta t$ for coming to rest during the collision is “infinitesimal” (i.e., essentially smaller than any finite time interval). The downward gravitational impulse during the collision, $mg\delta t$, is then assumed to be negligible compared to the collision impulses due to the normal and friction forces from the slope. This is assumed to be true even if the cylinder is initially rolling very slowly. In this case, the cylinder would necessarily lose contact with the floor for slope angles $\theta$ in the range $0 < \theta \leq 90^o$.

(ii) The time interval for coming to rest is a very small, but finite time interval. If the cylinder is initially rolling slowly enough, the collision impulses required to bring the cylinder to rest could be smaller than the impulse due to the weight. The cylinder would not necessarily lose contact with the floor for $0 < \theta \leq 90^o$.

In the first interpretation, I think only one of the multiple choice answers is correct. In the second interpretation, I believe all four answers are correct.

 

[haruspex]

OK. The key phrase is “immediately come to rest”. I see two ways to interpret this

(i) The time interval $\delta t$ for coming to rest during the collision is “infinitesimal” (i.e., essentially smaller than any finite time interval). The downward gravitational impulse during the collision, $mg\delta t$, is then assumed to be negligible compared to the collision impulses due to the normal and friction forces from the slope. This is assumed to be true even if the cylinder is initially rolling very slowly. In this case, the cylinder would necessarily lose contact with the floor for slope angles $\theta$ in the range $0 < \theta \leq 90^o$.

(ii) The time interval for coming to rest is a very small, but finite time interval. If the cylinder is initially rolling slowly enough, the collision impulses required to bring the cylinder to rest could be smaller than the impulse due to the weight. The cylinder would not necessarily lose contact with the floor for $0 < \theta \leq 90^o$.

In the first interpretation, I think only one of the multiple choice answers is correct. In the second interpretation, I believe all four answers are correct.

Well, these things can all get a bit murky when at the boundary of physical reality and idealised behaviour. But I would certainly have interpreted the question as your option (i).
Even with option (ii), you get tangled up with deformation of surfaces and what it means to remain in contact. The ground surface would, in reality, be slightly depressed by the cylinder. On impact with the ramp, that becomes depressed and the ground a little less depressed. For the question of losing contact, it then becomes a matter of how those two elastic changes compare.

 

[TSny]

Well, these things can all get a bit murky when at the boundary of physical reality and idealised behaviour. But I would certainly have interpreted the question as your option (i).
Even with option (ii), you get tangled up with deformation of surfaces and what it means to remain in contact. The ground surface would, in reality, be slightly depressed by the cylinder. On impact with the ramp, that becomes depressed and the ground a little less depressed. For the question of losing contact, it then becomes a matter of how those two elastic changes compare.

Yes, I tend towards the first interpretation also. But, it does bother me a little bit that even if the cylinder is rolling at an extremely slow speed when it hits the 90 degree wall it will leave the floor.

 

[Vibhor]

The downward gravitational impulse during the collision, $mg\delta t$, is then assumed to be negligible compared to the collision impulses due to the normal and friction forces from the slope.

the collision impulses required to bring the cylinder to rest could be smaller than the impulse due to the weight. T

Why did you focus only on impulse due to gravity and not taken impulse due to normal force from the ground into consideration during collision ? Why did you compare just the two impulses i.e impulse due to gravity with impulse due to friction from the inclined plane ( ignoring impulse due to normal from the ground) .

This is what I had thought ( in the 90° case) .

During the collision the net vertical impulse would be (Mg - N - f)Δt . But since Mg = N before the collision ,it would also balance each other during collision which means Mg = N during the collision . Thus , the net vertical impulse would be fΔt upwards . So,irrespective of whether Δt is finite or not , there would be a net impulse fΔt upwards .

Sorry , if I do not make sense .

 

[TSny]

If you have a block resting on a table, the normal force will balance the weight. But if you apply a little upward force to the block (not enough to lift it off the table) the normal force will decrease. The net vertical force on the block remains zero.

Similarly, in the 90 degree wall case, the upward friction from the wall might not be enough to lift the cylinder off the table. The net vertical impulse NΔt + fΔt - mgΔt could be zero.

But, if fΔt > mgΔt the net vertical impulse is clearly nonzero and the cylinder will have some vertical linear momentum after the collision.

 

[Vibhor]

Thank you very much TSny .Thanks haruspex .