Realistic Bus Cornering Speed: Physics Analysis of "Speed" Movie | 1994 Film

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Homework Help Overview

The discussion revolves around analyzing the physics of a bus cornering at high speed as depicted in the 1994 film "Speed." Participants are exploring whether the bus can safely navigate a 90-degree turn at an estimated speed of 81 kph, given specific parameters such as mass, coefficient of friction, and radius of the turn.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants calculate the centripetal force and frictional force acting on the bus, questioning whether these forces are sufficient to prevent the bus from rolling over during the turn.
  • There is a focus on understanding the relationship between torque and the forces involved, particularly how the geometry of the bus affects its stability during cornering.
  • Some participants inquire about the implications of the bus's dimensions and center of mass on its ability to maintain contact with the road.

Discussion Status

The discussion is active, with participants providing calculations and questioning the adequacy of the frictional force to prevent skidding or tipping. There is an exploration of the conditions under which the bus might tip over, but no consensus has been reached regarding the final outcome of the scenario.

Contextual Notes

Participants are working with specific values for the bus's mass, dimensions, and friction, while also considering the effects of geometry on stability. The discussion acknowledges the complexity of the problem due to the lack of detailed information about the bus's center of mass.

chris123
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Movie Physics "speed"

Homework Statement


In the 1994 movie "speed" the bus takes a 90 degree corner at an estimated speed of 81kph. in reality would the bus turn the corner safely or roll due to torque? (disregard the movement of people to one side of the bus.)

"m" of bus=6000kg
coeficient of friction tyres to road=0.9=u
"V" of bus=81kph=22.5ms/s/
"r"=15m
"g"=9.8

Homework Equations


Fcp=mv^2/r
Ff=mug
"T"torque=Fxr

The Attempt at a Solution


Fcp=mv^2/r=6000x22.5^2/15=202500
?
 
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chris123 said:

Homework Statement


In the 1994 movie "speed" the bus takes a 90 degree corner at an estimated speed of 81kph. in reality would the bus turn the corner safely or roll due to torque? (disregard the movement of people to one side of the bus.)

"m" of bus=6000kg
coeficient of friction tyres to road=0.9=u
"V" of bus=81kph=22.5ms/s/
"r"=15m
"g"=9.8

Homework Equations


Fcp=mv^2/r
Ff=mug
"T"torque=Fxr

The Attempt at a Solution


Fcp=mv^2/r=6000x22.5^2/15=202500
?


Welcome to PF.

How much frictional force does the bus make with the ground?
You've calculated the radial force OK.
 


Ff=ugm=0.9x22.5x6000=121500

hows that?

just need help with steps cheers,,,,
 


TORQUE

T=Fxr=1822500

soo how do i know if this alright or if it flips?
 


chris123 said:
TORQUE

T=Fxr=1822500

soo how do i know if this alright or if it flips?

The radial Force is not exactly the torque. Surely it will contribute to it.

The torque would be a moment about some pivot point, presumably the outer wheels if it was to flip on the turn.

Without further knowledge of the geometry of the bus, (its center of mass relative to the wheels) I'd think the only thing that would happen is that it would be sure to skid, because the frictional force is insufficient to maintain contact with the road.
 


LowlyPion said:
The radial Force is not exactly the torque. Surely it will contribute to it.

The torque would be a moment about some pivot point, presumably the outer wheels if it was to flip on the turn.

Without further knowledge of the geometry of the bus, (its center of mass relative to the wheels) I'd think the only thing that would happen is that it would be sure to skid, because the frictional force is insufficient to maintain contact with the road.

Length=12 metres height=3 metres width=2 metres

? does this help
 


chris123 said:
Length=12 metres height=3 metres width=2 metres

? does this help

If the frictional force is exceeded, then the bus should slide. Hence the force available to pivot about the outer wheel is going to be given by the force from friction. The forces then that balance the torque are the downward force of the weight through the center of mass times it's distance from the wheels compared to the frictional force times the height above the wheel. (Draw a diagram to satisfy yourself.)

If Friction Force X 1/2 height is greater than weight times 1/2 width then it will tip. Once tipped it should continue because the moment arm of the lateral force increases and the distance of the weight through the center of mass will move closer to the outside wheel.
 

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