How can the time taken for a loop to roll down an incline be calculated?

[Saitama]

Homework Statement

Homework Equations

The Attempt at a Solution

The force on the loop due to magnetic field in the direction normal to the incline is ##qvB##.
Balancing forces in the direction normal to the incline,
[tex]mg\cos\alpha=qvB+N[/tex]
where ##N## is the normal reaction due to incline and v is the velocity of CM of loop.
When the loop takes off, N=0 i.e
[tex]mg\cos\alpha=qvB[/tex]
Also, ##v=g\sin\alpha t##. Substituting this in the above expression and solving for t, I get ##t=1 sec## but this is wrong.

 

[tiny-tim]

Hi Pranav-Arora!

Also, ##v=g\sin\alpha t##.

Nope.

If you got this from F = ma, you left out the static friction force.

(were you really so optimistic as to think that you could solve this without using the rolling constratint? )

 

[spaderdabomb]

I would say if the loop has a mass, you will have to account for its rolling motion. This means use the rotational inertia for a thin loop!

 

[Saitama]

Nope.

If you got this from F = ma, you left out the static friction force.

Ah yes, completely missed it. But how am I supposed to calculate the friction force here? Do I have to go like this:
##mg\sin\alpha-f_R=ma## (Translatory motion)
##f_RR=I\beta=mR^2a/R## (Torque about CM, ##\beta## is the angular acceleration)
Solving the two equations:
##a=g\sin\alpha/2##
Hence, ##v=g\sin\alpha t/2##
I had,
##mg\cos\alpha=qvB##
Substituting v and solving for t, ##t=2 sec##, looks correct?

 

[spaderdabomb]

I think what tiny-tim meant was that you forgot about an effect of static friction. You should not have to calculate any friction directly in this problem. But, what the static friction does, is it causes the loop to roll, not slide. If there was no static friction, the loop would slide instead of roll, causing it to have 0 angular momentum. Since there is static friction though (implicitly implied), you must take into the moment of inertia of the loop.

 

[Saitama]

I think what tiny-tim meant was that you forgot about an effect of static friction. You should not have to calculate any friction directly in this problem. But, what the static friction does, is it causes the loop to roll, not slide. If there was no static friction, the loop would slide instead of roll, causing it to have 0 angular momentum. Since there is static friction though (implicitly implied), you must take into the moment of inertia of the loop.

I did take moment of inertia into account. Isn't my last post clear enough?

 

[spaderdabomb]

Ah ok missed it sorry

 

[spaderdabomb]

Hmmm I don't like what you did still though. There shouldn't be any friction in this problem. Is there any sliding occurring? If yes, then there should be friction in this problem. If no, then there should not be any friction. It only has the efffect of causing rotation. The frictional force you put in there kind of came out of nowhere (unless you can somehow justify it to me?)

 

[spaderdabomb]

Actually I am sorry, you did it fine. I just thought it was weird how you put the friction in there to begin with. I would have gone straight to the moment of inertia haha.

Also I should add, using forces may not be the best method. It will be much easier to use energy. We can assign a given height 'h' where the hoop takes off. If we call ground 0 potential energy, then energy due to gravity is -mgh. Our kinetic energy of the ball moving hoop is 1/2mv^2 and the energy from the rotations of the hoop is 1/2Iω^2 where ω is the angular velocity, a function of velocity (ω=v/r). Now use conservation of energy. But if what you did gets you the correct answer then that is fine.

 

[tiny-tim]

Hi Pranav-Arora!

(sorry, I've been out, enjoying myself! )

Ah yes, completely missed it. But how am I supposed to calculate the friction force here? Do I have to go like this:
##mg\sin\alpha-f_R=ma## (Translatory motion)
##f_RR=I\beta=mR^2a/R## (Torque about CM, ##\beta## is the angular acceleration)
Solving the two equations:
##a=g\sin\alpha/2##
Hence, ##v=g\sin\alpha t/2##

yes that's fine

btw, you can check it by using the "rolling mass", equal to I/r², in this case m (for a disc it's m/2, for a sphere it's 2m/5)

you add the "rolling mass" to actual mass giving 2m, then you use that in F = ma

(so for a disc v = gsinαt/(m + m/2) = 2gsinαt/3, and for a sphere = 5gsinαt/7) …

but i don't think the examiners would like that, so just use it for checking ​

you can also find v by τ = Iα about the point of contact (since it's the centre of rotation, that's ok), which is quicker since it avoids using (and eliminating) the static friction force

 

[Saitama]

Thank you for the help tiny-tim!

Hi Pranav-Arora!

(sorry, I've been out, enjoying myself! )


yes that's fine

btw, you can check it by using the "rolling mass", equal to I/r², in this case m (for a disc it's m/2, for a sphere it's 2m/5)

you add the "rolling mass" to actual mass giving 2m, then you use that in F = ma

(so for a disc v = gsinαt/(m + m/2) = 2gsinαt/3, and for a sphere = 5gsinαt/7) …

but i don't think the examiners would like that, so just use it for checking ​

you can also find v by τ = Iα about the point of contact (since it's the centre of rotation, that's ok), which is quicker since it avoids using (and eliminating) the static friction force

Great! Thank you for both the tips, these will surely be helpful ahead. :)