Problem 5: Motorcyclist revisited
In the group problems, you considered the forces acting on a motorcyclist driving in a vertical loop at constant speed, at the top and bottom of the loop. In this problem you’ll consider what happens at other points on the loop.
Point A is at the right-most part of the loop. Point B is at position that is 45° from point A, measured from the center of the loop as shown.
The radius of the loop is 2.5 m, and the combined mass of the
A. Draw a free-body diagram for point A. (Hint: The normal force is perpendicular to the loop, and the weight force is down. The motorcyclist is driving at constant speed, so there must be a third force acting on them. What is it?)
B. Determine the magnitude of all three forces on your FBD.
C. Draw a free-body diagram for point B. You may want to choose a different coordinate system than you did for point A.
D. Determine the magnitude of all three forces on your FBD.
A. To draw a free-body diagram for point A, we need to consider the forces acting on the motorcyclist at that point.
First, we have the weight force, which is always directed downward. This force can be represented by a vector pointing straight down.
Second, we have the normal force. At point A, the motorcyclist is pushed against the loop from the inside due to the circular motion. Therefore, the normal force is perpendicular to the loop and points towards the center of the loop.
Lastly, we need to consider the force that keeps the motorcyclist moving at a constant speed. This force is the net force and is directed towards the center of the loop. It is generally referred to as the centripetal force.
B. To determine the magnitudes of the three forces on the free-body diagram for point A, we can use Newton's second law, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration.
Since the motorcyclist is driving at a constant speed, the net force is zero (because there is no change in velocity). Therefore, the magnitudes of the three forces must balance each other.
The weight force can be calculated using the formula: weight force = mass * gravitational acceleration (usually represented by "g").
The magnitude of the normal force is equal to the weight force, since they must cancel each other out.
The magnitude of the centripetal force can be determined using the equation: centripetal force = mass * (velocity^2 / radius), where velocity represents the speed at which the motorcyclist is traveling.
C. To draw a free-body diagram for point B, we can choose a different coordinate system. Let's use a coordinate system where the x-axis is horizontal and points towards the center of the loop, and the y-axis is vertical and points upwards.
Similar to point A, we have the weight force pointing downward and the normal force pointing towards the center of the loop. However, since point B is at a 45° angle from point A, we also need to consider a tangential force that acts horizontally towards the center of the loop.
D. To determine the magnitudes of the three forces on the free-body diagram for point B, we can again use Newton's second law.
The weight force and the normal force will have the same magnitudes as in point A.
The magnitude of the tangential force can be determined using the equation: tangential force = mass * (velocity^2 / radius) * cos(angle), where angle represents the angle between points A and B (45° in this case).