Linear speed of car coasting on two tracks

In summary: The answer is that, yes, tangential velocity does remain unchanged. Because that is implicitly our reference frame/point as well? But isn't it just the distance between the point of rotation and train (treated as a point mass in this case) that we're concerned about in the equation L = r * mv...?In summary, the linear speed of a streetcar on a circular track is greater when coasting on a smaller circle, but the angular momentum remains the same.
  • #1
riddle_me_phys
10
0
This is another one of those concept problems where if I think about it one way, it makes sense, but if I think about it another way I don't reach the same conclusion...again, any clarifications in explaining this would be much appreciated.

A streetcar is freely coasting (no friction) around a large circular track. It is then switched
to a small circular track. When coasting on the smaller circle its linear speed is...


greater​
less​
unchanged​

2. Homework Equations : kinetic energy = 1/2mv^2



3. Attempt at solution

If I take the system to be the two tracks, Earth and the car, then I can consider this as a closed system, where energy is conserved. There is no change in potential energy in the system, and we'll assume that there is no friction/heat loss. Since in a closed system, no energy enters or leaves the system, and the mass of the car does not change, then we can conclude that the kinetic energy of the car must stay the same, and that the velocity must then be the same.

BUT...what if I try to look at this from a angular momentum approach? It is a closed system and there are no external torques. So then couldn't you liken this to a revolving ice skater problem? When the ice skater brings his/her arms in, the linear/rotational speed increases. So likewise, when the coasting car transitions to the smaller track, why doesn't it speed up like the ice skater but just stays at the same speed instead?
 
Physics news on Phys.org
  • #2
In order to transfer the car from a large to a small circle, is any energy input required?
 
  • #3
No energy input is required. The big/small tracks are tangential to each other. So the car is coasting on the bigger track, then at that connection point, naturally transitions into the smaller track. Somewhat similar to a figure 8.
 
  • #4
OK, so I think you're right about the energy.

But the angular momentum is a product of 2 variables (mass is constant). Could angular momentum still be conserved if the velocity is the same?
 
  • #5
paisiello2 said:
OK, so I think you're right about the energy.

But the angular momentum is a product of 2 variables (mass is constant). Could angular momentum still be conserved if the velocity is the same?

When you spiral in, the force from the track, which is perpendicular to your speed, is not in the direction of the middle of the spiral, so there is a torque that reduces your angular momentum.
 
  • #6
It is in the direction of the center of the small circle, so I don't see where the torque comes in. The frictionless tracks only cause a radial acceleration.
 
  • #7
paisiello2 said:
It is in the direction of the center of the small circle, so I don't see where the torque comes in. The frictionless tracks only cause a radial acceleration.

That's only true if the track is a circle. If the track gets closer to the center of the circle, then the direction of the acceleration, which is perpendicular to the track is not in the direction of the center, so there is a torque present.
 
  • #8
I thought the tracks were both a circle.

A torque must change the tangential velocity. I think the OP correctly reasoned that the tangential velocity remains unchanged.
 
  • #9
paisiello2 said:
I thought the tracks were both a circle.

A torque must change the tangential velocity. I think the OP correctly reasoned that the tangential velocity remains unchanged.

But how are you ever going from the larger to the smaller circle? There must be a section of track between them that's not a circle.
 
  • #10
If the circles are concentric, then you have to spiral in.

On the other hand...
If you suddenly are on a track on a different circle by way of a shared tangent, such as in a figure 8 , then you will likely be determining angular momentum about a completely different point. While the train travels around the first circle, its angular momentum w.r.t. the center of the second circle is not constant. Likewise, while the train travels around the second circle, its angular momentum w.r.t. the center of the first circle is not constant.
 
  • #11
The solution states that energy is conserved, so the answer is that, yes, tangential velocity does remain unchanged.

There is no external torque in the system, so I see that the rule of conservation of angular momentum it breaks must be not being about the same point (center of the first circle). But what is the specific reason that the point must be the same in order to say that momentum is conserved? Because that is implicitly our reference frame/point as well? But isn't it just the distance between the point of rotation and train (treated as a point mass in this case) that we're concerned about in the equation L = r * mv ?
 
  • #12
riddle_me_phys said:
The solution states that energy is conserved, so the answer is that, yes, tangential velocity does remain unchanged.

There is no external torque in the system, so I see that the rule of conservation of angular momentum it breaks must be not being about the same point (center of the first circle). But what is the specific reason that the point must be the same in order to say that momentum is conserved? Because that is implicitly our reference frame/point as well?
Define angular momentum of a point mass.
 
  • Like
Likes 1 person
  • #13
For an object with a fixed mass that is rotating about a fixed symmetry axis, the angular momentum is expressed as the product of the moment of inertia of the object and its angular velocity vector:

L = I w

Ok, so fixed axis. I guess I was overthinking it =)

Thanks SammyS!
 

Related to Linear speed of car coasting on two tracks

1. What is linear speed?

Linear speed, also known as tangential speed, is the rate at which an object travels along a straight path. It is the distance traveled per unit of time.

2. How is linear speed calculated?

Linear speed is calculated by dividing the distance traveled by the time it takes to travel that distance. The formula is: linear speed = distance / time. The unit of measurement for linear speed is typically meters per second (m/s).

3. How does the linear speed of a car coasting on two tracks differ from a car on one track?

The linear speed of a car coasting on two tracks will be the same as a car on one track, as long as the two tracks are parallel and the car is not accelerating. This is because the distance traveled and time taken to travel that distance will remain the same, regardless of the number of tracks.

4. Can the linear speed of a car coasting on two tracks change?

Yes, the linear speed of a car coasting on two tracks can change if the car accelerates or decelerates. This will affect the distance traveled and the time it takes to travel that distance, resulting in a change in linear speed.

5. How does the linear speed of a car coasting on two tracks affect its overall speed?

The linear speed of a car coasting on two tracks is just one component of its overall speed. The overall speed of a car also takes into account other factors such as acceleration, deceleration, and any external forces acting on the car. Therefore, the linear speed alone may not accurately represent the overall speed of the car.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
17K
  • Introductory Physics Homework Help
Replies
5
Views
19K
  • Introductory Physics Homework Help
Replies
4
Views
398
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top