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The force pointing out involved in a bussturn. 
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#1
May612, 04:18 PM

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Hi guys,
I have been reading on the board for a while, and I have been enjoying the information that i have gathered. Now I have a question for you, though. Have any of you seen the movie speed? In this movie they do a 90degree turn, and the buss flips due to the torque. They also tested this in mythbusters, I believe. I have been thinking about the calculations in order to find the maximum velocity through the turn, and they seem pretty straight forward. We have a torque due to the gravity (attacking the COM), and a torque due to the centripetal force (also attacking the COM). Both torques with respect to some axis at the outer wheels. When the torque from the centripetal force becomes larger than the torque from gravity, the buss will flip. These calculations seem to work, but HOW can I explain the part with the centripetal force? Because, it can not be the actually centripetal force, since this point inward and the buss obviously tilt the other way. Can this be explained with the centrifugal force? If so, how exactly, and also, why doesn't the real centripetal force not give a torque pointing inward the circle? Is this because the centripetal force is really just the frictional force between tire and road, and that this force does not attack in the COM? Thank you in advance :) 


#2
May612, 06:19 PM

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P: 26,148

hi jegerjon! welcome to pf!
centripetal force is not a separate force it is only an alternative name for the radially inward component of tension or friction or other force or forces on a body in the case of the bus, it is the friction force, which of course is irrelevant to tipping, since it has no torque about the tipping axis the easiest way to solve this is to use the rotating frame of the bus … in that frame, there are two relevant forces, mg vertically downward, and the centrifugal force mv^{2}/r horizontally outward 


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