# Cylinder rolling due to magnetic force

• archaic
In summary, the magnetic force is to the right and the net force on the centre of mass is ##F=F_B-f##. I know that ##F_B## is ##IdB##. I also know that ##rf=I_c\alpha=I_c\frac ar##, so that ##f=I_c\frac{a}{r^2}=\frac{ma}{2}## (because ##I_c=\frac12mr^2##), which gives me ##ma=IdB-\frac{ma}{2}\Leftrightarrow a=\frac{2IdB}{3m}\Leftrightarrow F=\frac{2IdB}{
archaic
Homework Statement
The apparatus in the figure is a distance ##h## from the ground. The cylinder of radius ##r## rolls without slipping. I am asked to find the total horizontal distance that will be travelled until the rod reaches the ground.
Relevant Equations
$$\mathbf{F_B}=I\mathbf L\times\mathbf B$$$$\tau_{net}=I\alpha$$
Hello!

The magnetic force is to the right. ##I_c## is the moment of inertia of the cylinder.
For the net force on the centre of mass, I have the frictional and magnetic forces ##F=F_B-f##. I know that ##F_B## is ##IdB##.
I also know that ##rf=I_c\alpha=I_c\frac ar##, so that ##f=I_c\frac{a}{r^2}=\frac{ma}{2}## (because ##I_c=\frac12mr^2##), which gives me ##ma=IdB-\frac{ma}{2}\Leftrightarrow a=\frac{2IdB}{3m}\Leftrightarrow F=\frac{2IdB}{3}##.
Using the work energy theorem, I get ##\frac12mv^2=\frac23IdBL\Leftrightarrow v=\sqrt{\frac{4IdBL}{3m}}##, which is the horizontal speed when the rod enters free-fall.
The vertical position when the cylinder touches the ground is such that ##0=h-\frac12g(\Delta t)^2\Leftrightarrow\Delta t=\sqrt{\frac{2h}{g}}##.
The total range would be ##L+v\Delta t=L+\sqrt{\frac{8IdBLh}{3mg}}##
Correct?
Thank you. :)

Last edited:
Your work looks correct to me. Instead of using the work-energy theorem, you could have used the kinematic equation ##v^2 = v_0^2 +2 a \Delta x## to find the speed at the start of free-fall.

In your application of the work-energy theorem, someone might ask why you didn't include the rotational KE of the rod. But, your work is correct.

etotheipi
TSny said:
In your application of the work-energy theorem, someone might ask why you didn't include the rotational KE of the rod. But, your work is correct.

If I'm not mistaken, this is because if the net force on the rod is ##\vec{F} = \sum \vec{F}_i = m\ddot{\vec{x}}_{\text{cm}}##, a quantity ##\tilde{W}## defined by\begin{align*}\tilde{W} = \int_{\vec{x}_{\text{cm},1}}^{\vec{x}_{\text{cm},2}} \vec{F} \cdot d\vec{x}_{\text{cm}} = \int_{\vec{x}_{\text{cm},1}}^{\vec{x}_{\text{cm},2}} m\ddot{\vec{x}}_{\text{cm}} \cdot d\vec{x}_{\text{cm}} &= \int_{t_1}^{t_2} m \ddot{\vec{x}}_{\text{cm}} \cdot \dot{\vec{x}}_{\text{cm}} dt \\ &= \int_{t_1}^{t_2} \frac{d}{dt} \left( \frac{1}{2} m \dot{\vec{x}}_{\text{cm}}^2 \right) dt \\ &= \Delta \left( \frac{1}{2} m \dot{\vec{x}}_{\text{cm}}^2 \right) = \Delta T_{\text{cm}} \end{align*}which is the change in 'kinetic energy of the centre of mass' only, and not the total change in kinetic energy ##\Delta T = \Delta T^* + \Delta T_{\text{cm}}## where, in this case, ##\Delta T^*## is the change in rotational kinetic energy [i.e. in the centre of mass frame].

The total work done by the Laplace force on all individual mass elements of the rod whilst the rod is rolling along the ramp will be the total change in kinetic energy ##W = \Delta T##, and this'll be different to the ##\tilde{W}## from before (which we obtained by multiplying the Laplace force + friction force by the displacement of the centre of mass of the rod). Because the total work done will actually be something like$$W = \sum_i \int_{\vec{x}_1}^{\vec{x}_2} \vec{F}_i \cdot d\vec{x}_i$$where ##\vec{F}_i## is the Laplace force on a small element ##m_i## of the rod somewhere, and ##d\vec{x}_i## the displacement of that small element. E.g the points of the rod currently in contact with the rails have ##\dot{\vec{x}}_i = \vec{0}## and the Laplace force has zero power on them, but the points of the rod currently at the top of the cycle have maximum speed, etc. It would be interesting to see if, if we did the double integral over mass elements in the rod, the total work done by the Laplace force would indeed give us the total change in kinetic energy [including rotational]!

Last edited by a moderator:
TSny
etotheipi said:
It would be interesting to see if, if we did the double integral over mass elements in the rod, the total work done by the Laplace force would indeed give us the total change in kinetic energy [including rotational]!
I don't think it would be very hard to show this. The net work done by all of the elementary Laplace forces (i.e., magnetic forces) acting on the mass elements will turn out to be just the total magnetic force times the distance that the center of mass moves: ##IdBL##. And you can check that this equals the total change in translational plus rotational KE of the rod.

etotheipi
@TSny thank you very much for your confirmation! I actually forgot about the rotational KE... Glad that it works.
From my textbook, ##IdB## is the force on the whole wire, so I basically thought of the whole cylinder as a point.
Thank you for mentioning that, @etotheipi. I have also asked myself the same question after having read TSny's post.

etotheipi

## 1. What is the principle behind cylinder rolling due to magnetic force?

The principle behind cylinder rolling due to magnetic force is based on the interaction between a magnetic field and a conductive material. When a magnetic field is applied to a conductive cylinder, it induces a current in the cylinder which creates a secondary magnetic field. This secondary magnetic field interacts with the original magnetic field, causing the cylinder to roll.

## 2. How does the strength of the magnetic field affect the rolling of the cylinder?

The strength of the magnetic field directly affects the rolling of the cylinder. The stronger the magnetic field, the greater the induced current and resulting secondary magnetic field, which leads to a stronger force and faster rolling of the cylinder.

## 3. What factors influence the speed of the cylinder rolling?

The speed of the cylinder rolling is influenced by several factors, including the strength of the magnetic field, the size and shape of the cylinder, the material of the cylinder, and the surface it is rolling on. Additionally, any external forces acting on the cylinder, such as friction, can also affect its speed.

## 4. Can the direction of the magnetic field affect the direction of the cylinder's rolling?

Yes, the direction of the magnetic field can affect the direction of the cylinder's rolling. The direction of the magnetic field determines the direction of the induced current in the cylinder, which in turn affects the direction of the secondary magnetic field and the resulting force on the cylinder.

## 5. Are there any real-world applications of cylinder rolling due to magnetic force?

Yes, there are several real-world applications of cylinder rolling due to magnetic force. One example is in magnetic levitation trains, where the magnetic force is used to propel the train forward. This principle is also used in some types of electric motors and generators to convert electrical energy into mechanical energy.

• Introductory Physics Homework Help
Replies
97
Views
3K
• Introductory Physics Homework Help
Replies
13
Views
1K
• Introductory Physics Homework Help
Replies
12
Views
349
• Introductory Physics Homework Help
Replies
39
Views
2K
• Introductory Physics Homework Help
Replies
13
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
407
• Introductory Physics Homework Help
Replies
23
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
506
• Introductory Physics Homework Help
Replies
18
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
604