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 radius of the circle is $10 \mathrm{~m}$. What are the magnitude and direction of the force of the boom on the car at the top of the circle if the car's speed there is (a) $5.0 \mathrm{~m} / \mathrm{s}$ and (b) $12 \mathrm{~m} / \mathrm{s}$ ?

Key Concepts

Free-Body Diagram Analysis

Creating a free-body diagram is crucial for visualizing and summing the forces acting on an object. In problems involving circular motion, it helps in identifying all the forces, such as gravitational force and tension or normal force, and their directions relative to the center of the circle.

Dynamics of Vertical Circular Motion

In vertical circular motion, the magnitude and direction of forces like tension and weight change depending on the position in the circle. At different points, such as the top of the circle, the interplay between gravitational force and the force exerted by the supporting structure determines whether the net force meets the requirement for centripetal acceleration.

Centripetal Acceleration

This concept refers to the acceleration that keeps an object moving in a circular path, directed toward the center of the circle. It is quantified by the formula a = v²/r, where v is the speed of the object and r is the radius of the circle. Understanding centripetal acceleration is fundamental to analyzing forces in circular motion.

Newton's Second Law in Circular Motion

Newton's Second Law, F = ma, is essential for connecting the net force acting on an object to its acceleration. In the context of circular motion, it helps determine the magnitude of the net inward (centripetal) force required to maintain the object’s motion along the circular path.

Transcript

00:01 In the question we have given that the car with the with the rider that is 5 kilo right so total weight that is equal to 5 kilo newton right and we have also given the radius of the circular are that is 10 meter so in the first question we need to find the force and the magnitude of force and direction of the form on the car when the when the it's speed at the top of the point is 5 m per second so see if this is the the circular roof and this is a top point so it's a speed if 5 minute per second then what will be the force and the direction on the car due to the room so basically we know that wet force will be downward right that force will be obviously downward and um mass in acceleration so mv square equals to uh weight force plus m a i guess fb equals nothing but force that will say mg will be downward and mbs square plus r will be also be the centrifugal that required in the center central direction so force due to boom on the car that will be nothing but mg plus m a mg plus mbs square r so that will be 5k that is 5 ,000 plus mv squared so mv mass is 5k newton well let's say g eg over here that is 10 so this will be mass into v square v is 5 over here 25 of on r10.

01:32 It will be devalued out over here and fb that comes out to be i guess 5 ,000 and plus of 1 2 ,500 right.

01:53 So i guess no, no, no, no, no.

01:55 Actually this force mg will be downward direction, right? and the centrifugal force, centrifugal force that will be in the outward direction, outward direction obviously.

02:05 So if we will draw one direction that means fb the net force due to this one on the car that will be just the n this one and minus of this one so minus it will be 301 point a one sorry three seven five zero newton right and we can say that 3 .75 3 .75 in the of 3 newton right and in the next scenario when velocity v is equal to 12 meter then when v is equal meter then fb equals to 5 000 and minus of 5 000 upon 10 into we have when 144 upon r and this will be 144 so 144 times of 50 so 144 into 144 into 50 so it will be 7 0000...

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