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when braking a car, why do the brakes heat up?
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My understanding is that if the car isn't slipping, that static friction at the point of contact between the tyre and the car is what decelerates the car, counteracting the torque which is being applied by the brakes.
This torque is due to kinetic friction (as the rotor has rotational velocity, and the brake pads are held in place) but since this kinetic friction is an internal force / not the net force, do the rotor/brake-pads still heat up? It appears as if they do in the extreme cases (F1 racing), even though the car doesn't appear to be slipping. I'm clearly missing something here.
newtonian-mechanics friction torque
asked Aug 15 at 16:04
roozbubu
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My understanding is that if the car isn't slipping, that static friction at the point of contact between the tyre and the car is what decelerates the car, counteracting the torque which is being applied by the brakes.
This torque is due to kinetic friction (as the rotor has rotational velocity, and the brake pads are held in place) but since this kinetic friction is an internal force / not the net force, do the rotor/brake-pads still heat up? It appears as if they do in the extreme cases (F1 racing), even though the car doesn't appear to be slipping. I'm clearly missing something here.
newtonian-mechanics friction torque
asked Aug 15 at 16:04
roozbubu
1044
"...contact between the tyre and the car is what decelerates the car, counteracting the torque which is being applied by the brakes" doesn't make much sense. Could you expand that a bit with a quick diagram or something? Alternatively, that system could be the cause for your misunderstanding.
– Chair
Aug 15 at 16:11
absolutely. I suspect this is where my picture is incomplete/incorrect, but this is correctly what I imagine the system to be: imgur.com/wVAuexU. The torque applied by the brakes is due to kinetic friction (brake pads/rotor at different velocities). The counteracting force (and thus, net force on the car) on the bottom in green is due to static friction, as the velocity at the point of contact between tyre/road is zero, so the tyre and the road are instantaneously at the same speed.
– roozbubu
Aug 15 at 16:25
Kinetic energy always causes heat loss. Regardless if it is internal or external. This is due to kinetic energy by its very nature always doing work $W=int F;dx$. This work must go somewhere - so it turns into heat.
– Steeven
Aug 15 at 18:30
add a comment |Â
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up vote
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down vote
favorite
My understanding is that if the car isn't slipping, that static friction at the point of contact between the tyre and the car is what decelerates the car, counteracting the torque which is being applied by the brakes.
This torque is due to kinetic friction (as the rotor has rotational velocity, and the brake pads are held in place) but since this kinetic friction is an internal force / not the net force, do the rotor/brake-pads still heat up? It appears as if they do in the extreme cases (F1 racing), even though the car doesn't appear to be slipping. I'm clearly missing something here.
newtonian-mechanics friction torque
asked Aug 15 at 16:04
roozbubu
1044
My understanding is that if the car isn't slipping, that static friction at the point of contact between the tyre and the car is what decelerates the car, counteracting the torque which is being applied by the brakes.
This torque is due to kinetic friction (as the rotor has rotational velocity, and the brake pads are held in place) but since this kinetic friction is an internal force / not the net force, do the rotor/brake-pads still heat up? It appears as if they do in the extreme cases (F1 racing), even though the car doesn't appear to be slipping. I'm clearly missing something here.
newtonian-mechanics friction torque
asked Aug 15 at 16:04
roozbubu
1044
edited Aug 15 at 16:34
asked Aug 15 at 16:04
roozbubu
1044
asked Aug 15 at 16:04
roozbubu
1044
asked Aug 15 at 16:04
roozbubu
1044
1044
"...contact between the tyre and the car is what decelerates the car, counteracting the torque which is being applied by the brakes" doesn't make much sense. Could you expand that a bit with a quick diagram or something? Alternatively, that system could be the cause for your misunderstanding.
– Chair
Aug 15 at 16:11
absolutely. I suspect this is where my picture is incomplete/incorrect, but this is correctly what I imagine the system to be: imgur.com/wVAuexU. The torque applied by the brakes is due to kinetic friction (brake pads/rotor at different velocities). The counteracting force (and thus, net force on the car) on the bottom in green is due to static friction, as the velocity at the point of contact between tyre/road is zero, so the tyre and the road are instantaneously at the same speed.
– roozbubu
Aug 15 at 16:25
Kinetic energy always causes heat loss. Regardless if it is internal or external. This is due to kinetic energy by its very nature always doing work $W=int F;dx$. This work must go somewhere - so it turns into heat.
– Steeven
Aug 15 at 18:30
add a comment |Â
"...contact between the tyre and the car is what decelerates the car, counteracting the torque which is being applied by the brakes" doesn't make much sense. Could you expand that a bit with a quick diagram or something? Alternatively, that system could be the cause for your misunderstanding.
– Chair
Aug 15 at 16:11
absolutely. I suspect this is where my picture is incomplete/incorrect, but this is correctly what I imagine the system to be: imgur.com/wVAuexU. The torque applied by the brakes is due to kinetic friction (brake pads/rotor at different velocities). The counteracting force (and thus, net force on the car) on the bottom in green is due to static friction, as the velocity at the point of contact between tyre/road is zero, so the tyre and the road are instantaneously at the same speed.
– roozbubu
Aug 15 at 16:25
Kinetic energy always causes heat loss. Regardless if it is internal or external. This is due to kinetic energy by its very nature always doing work $W=int F;dx$. This work must go somewhere - so it turns into heat.
– Steeven
Aug 15 at 18:30
"...contact between the tyre and the car is what decelerates the car, counteracting the torque which is being applied by the brakes" doesn't make much sense. Could you expand that a bit with a quick diagram or something? Alternatively, that system could be the cause for your misunderstanding.
– Chair
Aug 15 at 16:11
"...contact between the tyre and the car is what decelerates the car, counteracting the torque which is being applied by the brakes" doesn't make much sense. Could you expand that a bit with a quick diagram or something? Alternatively, that system could be the cause for your misunderstanding.
– Chair
Aug 15 at 16:11
absolutely. I suspect this is where my picture is incomplete/incorrect, but this is correctly what I imagine the system to be: imgur.com/wVAuexU. The torque applied by the brakes is due to kinetic friction (brake pads/rotor at different velocities). The counteracting force (and thus, net force on the car) on the bottom in green is due to static friction, as the velocity at the point of contact between tyre/road is zero, so the tyre and the road are instantaneously at the same speed.
– roozbubu
Aug 15 at 16:25
absolutely. I suspect this is where my picture is incomplete/incorrect, but this is correctly what I imagine the system to be: imgur.com/wVAuexU. The torque applied by the brakes is due to kinetic friction (brake pads/rotor at different velocities). The counteracting force (and thus, net force on the car) on the bottom in green is due to static friction, as the velocity at the point of contact between tyre/road is zero, so the tyre and the road are instantaneously at the same speed.
– roozbubu
Aug 15 at 16:25
Kinetic energy always causes heat loss. Regardless if it is internal or external. This is due to kinetic energy by its very nature always doing work $W=int F;dx$. This work must go somewhere - so it turns into heat.
– Steeven
Aug 15 at 18:30
Kinetic energy always causes heat loss. Regardless if it is internal or external. This is due to kinetic energy by its very nature always doing work $W=int F;dx$. This work must go somewhere - so it turns into heat.
– Steeven
Aug 15 at 18:30
add a comment |Â
1 Answer
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up vote
6
down vote
When braking, the majority of the kinetic energy of the car turns into heat at the brakes.
Here is how it works: when the car slows down its kinetic energy decreases and since we have conservation of energy, this energy needs to go some place.
The brake pads push against the discs. Due to friction between pad and the disc, friction energy is created. This is basically just heat in the discs and the pads.
The contact between tyre and road is really just a mechanism to transfer translational kinetic energy of the car into rotational energy at the wheels. But at first order, it's lossless. All the kinetic energy keeps the disc turning and rubbing against the pads, where it gets turned into heat.
So yes, the discs get hot (and very much so) when you are braking.
edited Aug 15 at 16:46
sammy gerbil
21.3k42355
answered Aug 15 at 16:39
Hilmar
56324
I suppose what I'm having a hard time with is seeing how both the static friction and kinetic friction can occur simultaneously? In the force diagram above, it appears as though the static friction is a direct reaction/response to the kinetic friction, rendering it an internal/not apparent force, no? How would both occur on the same body of mass (the tyre)?
– roozbubu
Aug 15 at 17:24
Ah, I think I've got it, thanks to: "The contact between tyre and road is really just a mechanism to transfer translational kinetic energy of the car into rotational energy at the wheels". The net force on the tyre is not entirely translational in the deceleration - the translational component come purely from the static friction at that point of contact, but that static friction does not counteract or cancel out the rotational deceleration due to kinetic friction. The two are linked by the fact that the tyre is not slipping, but this constraint is then a consequence of both of these forces?
– roozbubu
Aug 15 at 17:32
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
When braking, the majority of the kinetic energy of the car turns into heat at the brakes.
Here is how it works: when the car slows down its kinetic energy decreases and since we have conservation of energy, this energy needs to go some place.
The brake pads push against the discs. Due to friction between pad and the disc, friction energy is created. This is basically just heat in the discs and the pads.
The contact between tyre and road is really just a mechanism to transfer translational kinetic energy of the car into rotational energy at the wheels. But at first order, it's lossless. All the kinetic energy keeps the disc turning and rubbing against the pads, where it gets turned into heat.
So yes, the discs get hot (and very much so) when you are braking.
edited Aug 15 at 16:46
sammy gerbil
21.3k42355
answered Aug 15 at 16:39
Hilmar
56324
I suppose what I'm having a hard time with is seeing how both the static friction and kinetic friction can occur simultaneously? In the force diagram above, it appears as though the static friction is a direct reaction/response to the kinetic friction, rendering it an internal/not apparent force, no? How would both occur on the same body of mass (the tyre)?
– roozbubu
Aug 15 at 17:24
Ah, I think I've got it, thanks to: "The contact between tyre and road is really just a mechanism to transfer translational kinetic energy of the car into rotational energy at the wheels". The net force on the tyre is not entirely translational in the deceleration - the translational component come purely from the static friction at that point of contact, but that static friction does not counteract or cancel out the rotational deceleration due to kinetic friction. The two are linked by the fact that the tyre is not slipping, but this constraint is then a consequence of both of these forces?
– roozbubu
Aug 15 at 17:32
add a comment |Â
up vote
6
down vote
When braking, the majority of the kinetic energy of the car turns into heat at the brakes.
Here is how it works: when the car slows down its kinetic energy decreases and since we have conservation of energy, this energy needs to go some place.
The brake pads push against the discs. Due to friction between pad and the disc, friction energy is created. This is basically just heat in the discs and the pads.
The contact between tyre and road is really just a mechanism to transfer translational kinetic energy of the car into rotational energy at the wheels. But at first order, it's lossless. All the kinetic energy keeps the disc turning and rubbing against the pads, where it gets turned into heat.
So yes, the discs get hot (and very much so) when you are braking.
edited Aug 15 at 16:46
sammy gerbil
21.3k42355
answered Aug 15 at 16:39
Hilmar
56324
I suppose what I'm having a hard time with is seeing how both the static friction and kinetic friction can occur simultaneously? In the force diagram above, it appears as though the static friction is a direct reaction/response to the kinetic friction, rendering it an internal/not apparent force, no? How would both occur on the same body of mass (the tyre)?
– roozbubu
Aug 15 at 17:24
Ah, I think I've got it, thanks to: "The contact between tyre and road is really just a mechanism to transfer translational kinetic energy of the car into rotational energy at the wheels". The net force on the tyre is not entirely translational in the deceleration - the translational component come purely from the static friction at that point of contact, but that static friction does not counteract or cancel out the rotational deceleration due to kinetic friction. The two are linked by the fact that the tyre is not slipping, but this constraint is then a consequence of both of these forces?
– roozbubu
Aug 15 at 17:32
add a comment |Â
up vote
6
down vote
up vote
6
down vote
When braking, the majority of the kinetic energy of the car turns into heat at the brakes.
Here is how it works: when the car slows down its kinetic energy decreases and since we have conservation of energy, this energy needs to go some place.
The brake pads push against the discs. Due to friction between pad and the disc, friction energy is created. This is basically just heat in the discs and the pads.
The contact between tyre and road is really just a mechanism to transfer translational kinetic energy of the car into rotational energy at the wheels. But at first order, it's lossless. All the kinetic energy keeps the disc turning and rubbing against the pads, where it gets turned into heat.
So yes, the discs get hot (and very much so) when you are braking.
edited Aug 15 at 16:46
sammy gerbil
21.3k42355
answered Aug 15 at 16:39
Hilmar
56324
When braking, the majority of the kinetic energy of the car turns into heat at the brakes.
Here is how it works: when the car slows down its kinetic energy decreases and since we have conservation of energy, this energy needs to go some place.
The brake pads push against the discs. Due to friction between pad and the disc, friction energy is created. This is basically just heat in the discs and the pads.
The contact between tyre and road is really just a mechanism to transfer translational kinetic energy of the car into rotational energy at the wheels. But at first order, it's lossless. All the kinetic energy keeps the disc turning and rubbing against the pads, where it gets turned into heat.
So yes, the discs get hot (and very much so) when you are braking.
edited Aug 15 at 16:46
sammy gerbil
21.3k42355
answered Aug 15 at 16:39
Hilmar
56324
edited Aug 15 at 16:46
sammy gerbil
21.3k42355
edited Aug 15 at 16:46
sammy gerbil
21.3k42355
edited Aug 15 at 16:46
sammy gerbil
21.3k42355
21.3k42355
answered Aug 15 at 16:39
Hilmar
56324
answered Aug 15 at 16:39
Hilmar
56324
answered Aug 15 at 16:39
Hilmar
56324
56324
I suppose what I'm having a hard time with is seeing how both the static friction and kinetic friction can occur simultaneously? In the force diagram above, it appears as though the static friction is a direct reaction/response to the kinetic friction, rendering it an internal/not apparent force, no? How would both occur on the same body of mass (the tyre)?
– roozbubu
Aug 15 at 17:24
Ah, I think I've got it, thanks to: "The contact between tyre and road is really just a mechanism to transfer translational kinetic energy of the car into rotational energy at the wheels". The net force on the tyre is not entirely translational in the deceleration - the translational component come purely from the static friction at that point of contact, but that static friction does not counteract or cancel out the rotational deceleration due to kinetic friction. The two are linked by the fact that the tyre is not slipping, but this constraint is then a consequence of both of these forces?
– roozbubu
Aug 15 at 17:32
add a comment |Â
I suppose what I'm having a hard time with is seeing how both the static friction and kinetic friction can occur simultaneously? In the force diagram above, it appears as though the static friction is a direct reaction/response to the kinetic friction, rendering it an internal/not apparent force, no? How would both occur on the same body of mass (the tyre)?
– roozbubu
Aug 15 at 17:24
Ah, I think I've got it, thanks to: "The contact between tyre and road is really just a mechanism to transfer translational kinetic energy of the car into rotational energy at the wheels". The net force on the tyre is not entirely translational in the deceleration - the translational component come purely from the static friction at that point of contact, but that static friction does not counteract or cancel out the rotational deceleration due to kinetic friction. The two are linked by the fact that the tyre is not slipping, but this constraint is then a consequence of both of these forces?
– roozbubu
Aug 15 at 17:32
I suppose what I'm having a hard time with is seeing how both the static friction and kinetic friction can occur simultaneously? In the force diagram above, it appears as though the static friction is a direct reaction/response to the kinetic friction, rendering it an internal/not apparent force, no? How would both occur on the same body of mass (the tyre)?
– roozbubu
Aug 15 at 17:24
I suppose what I'm having a hard time with is seeing how both the static friction and kinetic friction can occur simultaneously? In the force diagram above, it appears as though the static friction is a direct reaction/response to the kinetic friction, rendering it an internal/not apparent force, no? How would both occur on the same body of mass (the tyre)?
– roozbubu
Aug 15 at 17:24
Ah, I think I've got it, thanks to: "The contact between tyre and road is really just a mechanism to transfer translational kinetic energy of the car into rotational energy at the wheels". The net force on the tyre is not entirely translational in the deceleration - the translational component come purely from the static friction at that point of contact, but that static friction does not counteract or cancel out the rotational deceleration due to kinetic friction. The two are linked by the fact that the tyre is not slipping, but this constraint is then a consequence of both of these forces?
– roozbubu
Aug 15 at 17:32
Ah, I think I've got it, thanks to: "The contact between tyre and road is really just a mechanism to transfer translational kinetic energy of the car into rotational energy at the wheels". The net force on the tyre is not entirely translational in the deceleration - the translational component come purely from the static friction at that point of contact, but that static friction does not counteract or cancel out the rotational deceleration due to kinetic friction. The two are linked by the fact that the tyre is not slipping, but this constraint is then a consequence of both of these forces?
– roozbubu
Aug 15 at 17:32
add a comment |Â
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"...contact between the tyre and the car is what decelerates the car, counteracting the torque which is being applied by the brakes" doesn't make much sense. Could you expand that a bit with a quick diagram or something? Alternatively, that system could be the cause for your misunderstanding.
– Chair
Aug 15 at 16:11
absolutely. I suspect this is where my picture is incomplete/incorrect, but this is correctly what I imagine the system to be: imgur.com/wVAuexU. The torque applied by the brakes is due to kinetic friction (brake pads/rotor at different velocities). The counteracting force (and thus, net force on the car) on the bottom in green is due to static friction, as the velocity at the point of contact between tyre/road is zero, so the tyre and the road are instantaneously at the same speed.
– roozbubu
Aug 15 at 16:25
Kinetic energy always causes heat loss. Regardless if it is internal or external. This is due to kinetic energy by its very nature always doing work $W=int F;dx$. This work must go somewhere - so it turns into heat.
– Steeven
Aug 15 at 18:30