Solving Banked Road Problem with Car Speeds of 48 km/h

  • Thread starter Thread starter Bernie Hunt
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SUMMARY

The discussion centers on calculating the banking angle for a car rounding a frictionless curve with a radius of 30 meters at a speed of 48 km/h. The initial calculation by the user Bernie yielded a banking angle of 82.7 degrees, which contradicts the textbook answer of 31 degrees. The error was identified as a units conversion mistake, emphasizing the importance of converting 48 km/h to meters per second. Additionally, another user raised a question about the forces acting on a car on a banked track, specifically regarding centripetal acceleration and the normal force.

PREREQUISITES
  • Understanding of centripetal acceleration and its formula: (mv^2)/r
  • Knowledge of forces acting on an object in circular motion, including normal force and gravitational force
  • Ability to convert units between kilometers per hour and meters per second
  • Familiarity with trigonometric functions, particularly tangent and sine
NEXT STEPS
  • Learn how to convert speeds from kilometers per hour to meters per second accurately
  • Study the derivation of the banking angle formula for circular motion
  • Explore the relationship between normal force and gravitational force on a banked curve
  • Practice solving problems involving banked curves with varying speeds and radii
USEFUL FOR

Students preparing for physics exams, particularly those focusing on mechanics and circular motion, as well as educators teaching these concepts in a classroom setting.

Bernie Hunt
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OK, last time I'll bother you guys today. I'm just having a tough time getting things to work out.

A curve of radius 30m is banked so that a car can round the curve at 48km/h even if the road is frictionless. Calculate the banking angle theta for these conditions.

I have;

tan(theta) = v^2 / gr

tan(theta) = 48^2 / (9.8 * 30)

tan(theta) = 7.8367

theta = 82.7 deg

The book has 31 deg.

Any ideas?

Thanks,
Bernie
 
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48km/h is a dangerous thing to plug into an equation like that. As a habbit always plug the units in with their numbers
 
You will have to convert 48 km/h to m/s. Or, convert 30m to km and 9.8 m/s^2 to km/hr^2.
 
Argh, Thud, thud, thud ...
(The sound of beating my head on the desk again.(

That's the second time I made a units mistake last night. My montra for today will be "Check the units!"

Bernie
 
Help!

Okay, so I am doing a similar problem involving a car driving on a banked, circular track (theta=31degrees). I know that to find the centripetal acceleration, I am supposed to say that (mv^2)/r = nsin(theta). Then, I have to solve for n by saying that ncos(theta)=mg. However, I am confused... why can't n=mgcos(theta). My understanding is that two forces are equal in magnitude if the object doesn't move in either direction. The car doesn't move into the road or out of the road... or does it?? please help! I have a test on monday.
 

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