Rotating cylinder rolling without slipping in a B field

In summary, the electric field is causing the cylinder to move down the incline. The force due to the electric field is acting down the slope, so the FBD looks like: The two arrows pointing towards the right represent the force due to the field and weight of the cylinder. The equation for the translation down the incline is: I need to find the linear acceleration and the direction of the magnetic force.
  • #1
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Homework Statement
Cylinder length 1.2m, mass =0.1kg, radius 0.5cm is released from rest and roll down without slipping down 2 rails connected to the battery of 10V. Rails at angle of 40 degrees. B = 0.01T. The total resistance of rails and cylinder is 250 ohms. Calculate V after it has rolled down 0.45m
Relevant Equations
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1569851680557.png

Firstly, I need to determine what the electric field is causing.
Using left hand rule, the force due to the field is acting down the slope.
Hence my FBD looks like:

1569851782386.png

Where the two arrows pointing towards the right represent the force due to the field and weight of the cylinder.
Since :

##\epsilon = blv##
##I =\frac{\epsilon}{r} = \frac{blv}{R}##
Is it correct to say that:
##ma = mgsin\theta + F_{b} = mgsin\theta + bIl## ?

If so, what do I do next?

Rolling down 0.45m , does that mean I need to compute the time it takes to roll down using Pythagoras theorem? Or am I overthinking here?

Thank you
 
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  • #2
jisbon said:
Firstly, I need to determine what the electric field is causing.
Using left hand rule, the force due to the field is acting down the slope.
Hence my FBD looks like:

View attachment 250424
Where the two arrows pointing towards the right represent the force due to the field and weight of the cylinder.
Please draw a better FBD and label all the forces. Do not include both the components of a force and the resultant.
jisbon said:
Since :

##\epsilon = blv##
Doesn't the fact that the field is not at a right angle relative to the velocity make a difference?
jisbon said:
##I =\frac{\epsilon}{r} = \frac{blv}{R}##
Is it correct to say that:
##ma = mgsin\theta + F_{b} = mgsin\theta + bIl## ?
No, it is not correct. If the cylinder rolls without slipping, there is an additional force acting on it. What is it?
jisbon said:
If so, what do I do next?
You need to write an additional equation expressing Newton's 2nd law for rotations and combine that with the correct equation for the translation down the incline to find the linear acceleration. You also need to figure out the correct direction for the magnetic force on the cylinder.

I note that is problem is ill posed. As the cylinder rolls down, the amount of rail that is part of the circuit becomes shorter and hence the contribution of the rail to the resistance decreases. Usually, the rails have zero resistance and either the cylinder has resistance R or the cylider also has zero resistance but the rails are bridged with resistor R.
 
  • #3
Will the magnetic force on the cylinder have any appreciable effect on the motion of the cylinder? Compare the numerical values of ##mg## and ##bIl## for a current due to the 10 Volts and 250 Ohms.
 
  • #4
kuruman said:
Please draw a better FBD and label all the forces. Do not include both the components of a force and the resultant.
Do you mean by this?

1569896721068.png

kuruman said:
Doesn't the fact that the field is not at a right angle relative to the velocity make a difference?

In this case is the EMF:

##\epsilon = BLVsin\theta##
whereby ##\theta## = 90-40 =50 degrees?

kuruman said:
No, it is not correct. If the cylinder rolls without slipping, there is an additional force acting on it. What is it?

That will be friction, as stated in my updated FBD.

kuruman said:
You need to write an additional equation expressing Newton's 2nd law for rotations and combine that with the correct equation for the translation down the incline to find the linear acceleration. You also need to figure out the correct direction for the magnetic force on the cylinder.

I note that is problem is ill posed. As the cylinder rolls down, the amount of rail that is part of the circuit becomes shorter and hence the contribution of the rail to the resistance decreases. Usually, the rails have zero resistance and either the cylinder has resistance R or the cylider also has zero resistance but the rails are bridged with resistor R.

So combined with my updated existing equation where:

##ma = mgsin\theta + F_{b} - f_{s} = mgsin\theta + BIL -\mu_{k} (mgcos\theta)##

Finding an equation for rotation:

##\tau = I\alpha = \mu_{k}(mgcos\theta)(0.5cm)##

So am I supposed to find out the rotational acceleration? If so, what is the inertia?
 
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  • #5
jisbon said:
Do you mean by this?

View attachment 250465
Yes, something like this. The direction of Fb is incorrect. What do you need to do to find the correct direction?
jisbon said:
In this case is the EMF:
##\epsilon = BLVsin\theta##
whereby ##\theta## = 90-40 =50 degrees?
In a steady, uniform field, the magnitude of the induced emf is given by ##\epsilon=\dfrac{d}{dt}(\vec B\cdot\hat n~A)##, where ##\hat n## is the normal to the rectangular loop the area ##A## of the loop that is shrinking. Figure it out from there.
jisbon said:
That will be friction, as stated in my updated FBD.
Yes, there is friction, but what kind, static or kinetic? If kinetic, it is opposite to the velocity; if static, it is in whatever direction is appropriate to provide the observed acceleration.
jisbon said:
So combined with my updated existing equation where:
##ma = mgsin\theta + F_{b} - f_{s} = mgsin\theta + BIL -\mu_{k} (mgcos\theta)##
Even though you use ##f_s## to denote static friction you set it equal to the expression of kinetic friction ##\mu_k N##. You cannot do this.
jisbon said:
Finding an equation for rotation:
##\tau = I\alpha = \mu_{k}(mgcos\theta)(0.5cm)##
See comment about friction above.
jisbon said:
So am I supposed to find out the rotational acceleration? If so, what is the inertia?
Yes, find the rotational acceleration and relate it to the linear acceleration. For this particular problem, it is easier to calculate torques about the point of contact. Even though the number of torques is more than one about that point, the torque provided by friction is zero so you don't need to know what it is and the torque equation is sufficient to give you the acceleration. You can look up the moment of inertia of a cylinder about its axis on the internet. Should you choose to write the torque equation about the point of contact as I suggested, don't forget to use the parallel axis theorem to find the moment of inertia about that point.

A final note: To avoid confusion, don't use ##I## for both current and moment of inertia. By the way, what is an expression for the current through the cylinder?
 
  • #6
kuruman said:
Yes, something like this. The direction of Fb is incorrect. What do you need to do to find the correct direction?
Oh, I figured that my left-hand rule was wrong because I misread how current is travelled.

I figured out that the force caused by magnetic field is: F=BIL(sin50) since the angle between current and magnetic field is 90-40.

Upon further asking my teacher for hints, he stated that with this force I can simply find the acceleration by diving by mass since F=ma.

The question is what about friction and mgsin theta? Why is it applicable to ignore in this case?

Thanks
 
  • #7
jisbon said:
Oh, I figured that my left-hand rule was wrong because I misread how current is travelled.
It's a right hand rule. The left hand rule gives you the opposite direction.
jisbon said:
I figured out that the force caused by magnetic field is: F=BIL(sin50) since the angle between current and magnetic field is 90-40.
You figured out incorrectly. Look at your drawing. The fact that the cylinder lies on an incline is irrelevant.
jisbon said:
Upon further asking my teacher for hints, he stated that with this force I can simply find the acceleration by diving by mass since F=ma.
Your teacher's suggestion is to the point but not very useful. You can always find the acceleration by dividing the net force Fnet by the mass. The real question is how to find an expression for Fnet. To do that (a) draw a complete free body diagram; (b) find expressions for all the forces in the diagram in terms of symbols representing the given quantities and (c) add all these forces as vectors to find the net force.
jisbon said:
The question is what about friction and mgsin theta? Why is it applicable to ignore in this case?
Friction is one of the forces acting on the cylinder and ##mg\sin\theta## is the component of the weight parallel to the incline. Neither is (or should be) ignored.

The two equations you must express in terms of the given quantities symbolically are$$\vec F_{net}=m\vec a_{cm}~;~~\vec {\tau}_{net}=I\vec {\alpha}$$Because the cylinder rolls without slipping, you can relate the magnitude of the acceleration of the center of mass to the magnitude of the angular acceleration by ## a_{cm}=\alpha R##, where ##R## is the radius of the cylinder, not the resistance.

Perhaps you should first solve this problem in zero magnetic field with the cylinder just rolling down the incline due to the effect of gravity alone. That will show you how the two equations can be brought together to get the acceleration. Once you understand that, then you can add the magnetic field. As @TSny suggested in #3, you may find that given the numbers in this problem, the effect of having the field on is just as negligible as air resistance. IMO the pedagogical value of this problem lies in solving it symbolically as given.
 
  • #8
kuruman said:
As @TSny suggested in #3, you may find that given the numbers in this problem, the effect of having the field on is just as negligible as air resistance. IMO the pedagogical value of this problem lies in solving it symbolically as given.
Is that why the resulting ##F_{net}## is only due to the force made by the field as a result?
 
  • #9
kuruman said:
The fact that the cylinder lies on an incline is irrelevant.
1570066567846.png

Isn't the red angle supposed to be angle between the field and the velocity (hence 90-40)?
 
  • #10
Yes, the red angle is 90-40 degrees, but that's not the issue. The B-field is straight up in the figure.You are looking for the angle between the B-field and the direction of the current through the cylinder. What is the direction of the current in the figure? Hint: It is not along the incline.

 

1. What is a rotating cylinder rolling without slipping in a B field?

A rotating cylinder rolling without slipping in a B field is a physical phenomenon where a cylinder is placed on a surface and rotates while maintaining contact with the surface, while also experiencing a magnetic field (B field) perpendicular to its motion.

2. What causes a rotating cylinder to roll without slipping in a B field?

The rolling motion of the cylinder is caused by the interaction between the magnetic field and the induced electric current in the cylinder. This current creates a magnetic field that repels the original field, causing the cylinder to rotate in the opposite direction.

3. How does the speed of the cylinder affect its rotation in a B field?

The speed of the cylinder does not affect its rotation in a B field, as long as it is rolling without slipping. This is because the induced current and resulting magnetic field are dependent on the rotational speed of the cylinder, not its linear speed.

4. Can a rotating cylinder in a B field change direction?

Yes, a rotating cylinder in a B field can change direction. This is due to the fact that the induced current and magnetic field depend on the direction of rotation of the cylinder. If the direction of rotation changes, the direction of the induced current and magnetic field will also change, causing the cylinder to change direction.

5. What are some real-life applications of a rotating cylinder rolling without slipping in a B field?

This phenomenon is commonly seen in electric motors and generators, where a rotating coil experiences a magnetic field and converts mechanical energy into electrical energy or vice versa. It is also used in some forms of transportation, such as maglev trains, where the train is levitated and propelled by the interaction between a rotating magnetic field and a stationary magnetic field on the track.

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