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

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SUMMARY

The analysis of the bus cornering speed in the 1994 film "Speed" reveals that at an estimated speed of 81 kph (22.5 m/s), the bus, weighing 6000 kg, would likely skid rather than roll due to insufficient frictional force. The coefficient of friction between the tires and the road is 0.9, resulting in a frictional force of 121,500 N. The calculated torque about the outer wheels is 1,822,500 Nm, indicating that without proper geometry of the bus's center of mass, it is more probable that the bus would slide rather than tip over during the turn.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with torque and its calculations
  • Knowledge of friction coefficients and their implications
  • Basic geometry related to center of mass and pivot points
NEXT STEPS
  • Study the principles of angular momentum and its effects on vehicle stability
  • Learn about the dynamics of vehicle cornering and weight transfer
  • Explore advanced friction models in vehicle dynamics
  • Investigate the impact of vehicle design on cornering performance
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Physics students, automotive engineers, and anyone interested in vehicle dynamics and safety analysis will benefit from this discussion.

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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