Asked by anonymous
The question is:
Sometimes, road surfaces have banked curves. Use an FBD to explain how this helps cars
to make turns more safely.
My answer is:
Road surfaces have banked curves to allow for more centripetal force, which then allows for vehicle to remain in the turn and not sliding away at higher speeds. The banked curves create more centripetal force because of the angle they make with the ground by using the horizontal component of the normal force from the curved ramp. Compared to a flat surface, less friction force is needed on a curved road because the horizontal component of the normal force adds more centripetal force to the force of friction. In the flat surface road, turning will only utilize the friction force.
is answer correct?
Sometimes, road surfaces have banked curves. Use an FBD to explain how this helps cars
to make turns more safely.
My answer is:
Road surfaces have banked curves to allow for more centripetal force, which then allows for vehicle to remain in the turn and not sliding away at higher speeds. The banked curves create more centripetal force because of the angle they make with the ground by using the horizontal component of the normal force from the curved ramp. Compared to a flat surface, less friction force is needed on a curved road because the horizontal component of the normal force adds more centripetal force to the force of friction. In the flat surface road, turning will only utilize the friction force.
is answer correct?
Answers
Answered by
Damon
No it is not correct.
The centripetal force has to be m v^2/R no matter how you slope the road.
It the road is not sloped, the frictional force must be great enough to produce that force. mu m g = m v^2/R
Sloping the road allows the road to also produce a component of the normal force inward helping the friction force counter that centripetal effect. Imagine if you turned the road straight up like a vertical cylinder. Then the normal force would be straight toward the center. (Of course then the frictional force mu m v^2/R would have to equal mg. This used to be a popular thing at state fairs with a rotating cylinder. People would stand against the wall and friction would hold them as the floor dropped away.
The centripetal force has to be m v^2/R no matter how you slope the road.
It the road is not sloped, the frictional force must be great enough to produce that force. mu m g = m v^2/R
Sloping the road allows the road to also produce a component of the normal force inward helping the friction force counter that centripetal effect. Imagine if you turned the road straight up like a vertical cylinder. Then the normal force would be straight toward the center. (Of course then the frictional force mu m v^2/R would have to equal mg. This used to be a popular thing at state fairs with a rotating cylinder. People would stand against the wall and friction would hold them as the floor dropped away.
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