# Speed of the rod as it leaves the rails?

• nawand
In summary, the rod of mass 0.720 kg and radius 6.00 cm rests on two parallel rails that are the d=12.0 cm apart and L=45.0 cm long. The rod carries a current of I=48.0 A in the direction shown and rolls along the rails without sleeping. A uniform magnetic field of magnitude 0240 T is directed perpendicular to the rod and the rails. If it starts from rest, what is the speed of the rod as it leaves the rails?The rod gains energy from the rest, so begin kinetic energy and rotational energy are zero. The only source of energy rod gains is magnetic filed. I= (mr2)/2 K=(mv2)/2
nawand

## Homework Statement

A rod of mass 0.720 kg and radius 6.00 cm rests on two parallel rails fig.1 that are the d=12.0 cm apart and L=45.0 cm long. The rod carries a current of I=48.0 A in the direction shown and rolls along the rails without sleeping. A uniform magnetic field of magnitude 0240 T is directed perpendicular to the rod and the rails. If it starts from rest, what is the speed of the rod as it leaves the rails?

## Homework Equations

0+0+Fb *L*cos θ = (mv2)/2 +(Iω2)/2

## The Attempt at a Solution

The rod gains energy from the rest, so begin kinetic energy and rotational energy are zero. The only source of energy rod gains is magnetic filed.
I = (mr2)/2
K=(mv2)/2
w=v/r
That is right side of equations becomes:
(mv2)/2 + 1/2 (mv2)+1/2(1/2 mr2) * (v2/r2)

Fb * L = 3/4 mv2
Where
Fb = I * d * B
etc
v=1.07 m/s

Because the force move the rod parallel to the rails along x axis, the dot product will be:
E=Fb * L * cos θ = Fb*L*1=FbL
The question is how do we get the value of energy exerts magnetic field on the rod?
That is force time Length of rails (Fb times L)?

For the electric field is E = qFe

#### Attachments

• Fig.1.png
2.6 KB · Views: 762
nawand said:
(mv2)/2 + 1/2 (mv2)+1/2(1/2 mr2) * (v2/r2)
It looks like you are adding the kinetic energy twice here (highlighted in bold).
The line afterwards is right again.
nawand said:
The question is how do we get the value of energy exerts magnetic field on the rod?
That is force time Length of rails (Fb times L)?
Sure.

I don't see an electric field.

Slipping, not sleeping, by the way.

nawand
Great!.

Indeed I have added it twice mv2/2 (This is my type fault).
• I was wondering about the energy was added in the left side Fb times L and which is likely the energy contribution of magnetic field.
• "Sleeping" = "Slipping"
Thanks a lot!

## 1. What is the speed of the rod as it leaves the rails?

The speed of the rod as it leaves the rails depends on various factors such as the initial velocity of the rod, the angle at which it is launched, and the amount of friction present. It can be calculated using the laws of motion and taking into account these factors.

## 2. How does the speed of the rod change as it leaves the rails?

The speed of the rod changes as it leaves the rails due to the forces acting on it. Initially, there is a decrease in speed as the rod experiences friction from the rails. However, as it gains momentum, the speed increases until it reaches its maximum velocity.

## 3. Can the speed of the rod be controlled?

Yes, the speed of the rod can be controlled to some extent by adjusting the initial velocity and the angle at which it is launched. However, certain external factors such as air resistance and surface friction may also affect the final speed of the rod.

## 4. How does the speed of the rod leaving the rails affect its trajectory?

The speed of the rod leaving the rails directly affects its trajectory. A higher speed will result in a longer and flatter trajectory, while a lower speed will result in a shorter and steeper trajectory. This is because the speed determines the distance the rod can cover before gravity pulls it down.

## 5. Why is it important to understand the speed of the rod as it leaves the rails?

Understanding the speed of the rod as it leaves the rails is crucial in predicting its trajectory and ensuring its safe and accurate launch. It is also essential in various applications, such as in ballistics and sports, where the speed of an object greatly affects its performance.

• Introductory Physics Homework Help
Replies
25
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
730
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
13
Views
1K
• Introductory Physics Homework Help
Replies
31
Views
3K
• Introductory Physics Homework Help
Replies
8
Views
3K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K