You need to make a sharp turn on a flat road, making a radius of curvature of 15

Question

You need to make a sharp turn on a flat road, making a radius of curvature of 15 meters. How does the required force of static friction between your tires compare if you make the turn at 30 mph vs. 60 mph? Select one: a. The force of friction needs to be twice as large. b. None of the above c. The force of friction needs to be four times as large. d. the force of friction is the same for both speeds since the radius of curvature is the same.
You need to make a sharp turn on a flat road, making a radius of curvature of 15 meters. How does the required force of static friction between your tires compare if you make the turn at 30 mph vs. 60 mph? Select one: a. The force of friction needs to be twice as large. b. None of the above c. The force of friction needs to be four times as large. d. the force of friction is the same for both speeds since the radius of curvature is the same. You need to make a sharp turn on a flat road, making a radius of curvature of 15 meters. How does the required force of static friction between your tires compare if you make the turn at 30 mph vs. 60 mph? Select one: a. The force of friction needs to be twice as large. b. None of the above c. The force of friction needs to be four times as large. d. the force of friction is the same for both speeds since the radius of curvature is the same. You need to make a sharp turn on a flat road, making a radius of curvature of 15 meters. How does the required force of static friction between your tires compare if you make the turn at 30 mph vs. 60 mph? Select one: a. The force of friction needs to be twice as large. b. None of the above c. The force of friction needs to be four times as large. d. the force of friction is the same for both speeds since the radius of curvature is the same. You need to make a sharp turn on a flat road, making a radius of curvature of 15 meters. How does the required force of static friction between your tires compare if you make the turn at 30 mph vs. 60 mph? Select one: a. The force of friction needs to be twice as large. b. None of the above c. The force of friction needs to be four times as large. d. the force of friction is the same for both speeds since the radius of curvature is the same. Select one: a. The force of friction needs to be twice as large. b. None of the above c. The force of friction needs to be four times as large. d. the force of friction is the same for both speeds since the radius of curvature is the same. a. The force of friction needs to be twice as large. b. None of the above c. The force of friction needs to be four times as large. d. the force of friction is the same for both speeds since the radius of curvature is the same.

Explanation / Answer

This is a problem related to rotational motion. In this case, the centripetal force is provided by the static friction.

Ff = Fc = (mv^2)/r

Where m and r are constant in this situation.

Doubling the velocity, therefore quadruples the centripetal force necessary for circular motion, which further means four times the force of friction.

So, the answer is C) The force of friction needs to be four times as large.

Hope that helps, cheers=) !

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