An amusement park ride consists of a car moving in a...

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 5.0 $\mathrm{kN},$ and the circle's radius is 10 $\mathrm{m} .$ At the top of the circle, what are the (a) magnitude $F_{B}$ and (b) direction $(u p$ or down $)$ of the force on the car from the boom if the car's speed is $v=5.0 \mathrm{m} / \mathrm{s}$ ? What are $(\mathrm{c}) F_{B}$ and (d) the direction if $v=12 \mathrm{m} / \mathrm{s} ?$

Key Concepts

Free-Body Diagrams

Free-body diagrams are essential tools that diagrammatically represent all the forces acting on an object. In the context of vertical circular motion, drawing a free-body diagram helps isolate forces such as weight and the force from a supporting structure, simplifying the application of Newton's Second Law to analyze the motion and determine unknown forces.

Force Analysis in Vertical Circular Motion

When an object moves in a vertical circle, the interplay between gravitational force and other forces (like tension or the force from a boom) must be considered. At the top of the circle, both weight (acting downward) and the applied force contribute to the net force required for centripetal acceleration. Analyzing these forces correctly determines the magnitude and direction of the forces acting on the object.

Newton's Second Law in Circular Motion

Newton’s Second Law, F = ma, underpins the analysis of circular motion by relating the net radial force acting on an object to its mass and centripetal acceleration. At any point in the motion, especially at critical points like the top of a vertical circle, the sum of the forces directed towards the center must equal m(v²/r), which aids in solving for forces such as the one from the boom in the amusement ride.

Centripetal Acceleration

In circular motion, an object requires an acceleration directed toward the center of the circle, known as centripetal acceleration. This acceleration is mathematically expressed as a = v²/r, where v is the object's speed and r is the radius of the circle. It is the central concept in determining the net force needed to maintain circular motion.

Transcript

00:01 So we can say this situation is very similar to figure 610.

00:10 And here, but instead of a downward normal force, we are dealing with the force of the boom on the car, which is capable of pointing in any direction.

00:21 So we can assume it, we will assume for the force of the boom to be upward as we apply newton's second law to the car.

00:31 Again we have a total weight of 5 ,000 newtons but for now we'll choose it to be upward and then if we're getting a negative answer we know that we have chosen the wrong direction so we can say that force applying new and second law we can say the force of the boom minus w will equal m a where the mass is equaling the weight divided by g and the acceleration would be the centripetal acceleration and this would equal negative v squared over r so we're only saying that the centripetal acceleration is negative because the centripetal acceleration is downward.

01:12 We're going to, downward, which we have chosen to be negative.

01:15 So we can say that upwards is positive.

01:22 And if we choose that, this means that the centripetal force would be negative because it is pointing downwards.

01:29 So we can say for part a, if r is equaling 10 meters, v is equaling 5 .0 meters per second.

01:39 We can then solve for the force of the boom, equaling the mass negative mv squared over r plus the weight, where this would simply be equal to mg minus mv squared over r.

02:07 And so essentially force of the boom would be equal to m times g minus v squared over r and plugging in these plugging in these uh essentially rather you know we i uh unfortunately we don't have a mass um so here we should actually say that this is going to equal to the weight over g plug in this for the mass times g minus v squared over r so this would allow us to actually plug in our knowns and again the weight is 5 ,000 newtons so the force of the boom would then be equal to 5 ,000 newt's 9 .8 percent applied 9 .8 meter per second squared 5 .0 meters per second quantity squared divided by r of 10 meters close parentheses and we find that here the force of the boom would be equal to approximately 3 .7 times 10 to the third newton's.

03:27 Now, this would be your answer for part a, the magnitude of the force of the boom.

03:32 For part b, they simply want the direction in part a...

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