A small car with mass 0.800 0.800 kg travels at constant speed on the inside of a track that is a vertical circle with radius 5.00 5.00 m (Fig. E 5.45 5.45 ). If the normal force exerted by the track on the car when it is at the top of the track (point B B ) is 6.00 6.00 N, what is the normal force on the car when it is at the bottom of the track (point A A )?

Welcome back everybody, we have a skateboarder that is going back in forth on this half pipe right here. And we are told a couple different things about the skateboarder, we are told that his mass is 55 kg and we are also told that the radius of curvature of this ramp is five meters. Were also told that at the bottom here of the ramp he reaches a max speed of eight m per second. And we are tasked with finding what the force is of the ramp acting on the skater. So I'm just gonna call that force up. Well the skaters not floating away and the ramp is not collapsing below him. So we know that force up is going to be equal to the magnitude of the summation of all forces acting downward or from the skater onto the ramp. What are those forces? Well, we are going to have the force due to gravity which is just going to be the master skater times gravitational acceleration, since he is also following centripetal motion here, we know that the skater is going to be pushing down on the ramp with this force right here of mass, times his velocity squared times the radius of curvature. Now, if we sum these two forces together and then take the absolute value, we are going to find the force of our upward force. Let's go ahead and plug in those values here. Force of the upward force is going to be the absolute value of well, we have our mass, which is 55 times our acceleration due to gravity, which is 9.8 plus our mass once again times our speed squared all over our radius, which when you plug this into your calculator, you get that. The upward force Is 1, Newtons, which corresponds to answer choice. C. Thank you guys so much for watching. Hope this video helped. We will see you all in the next one.

8. Centripetal Forces & Gravitation

Centripetal Forces

Physics

8. Centripetal Forces & Gravitation

Centripetal Forces

2:18 minutes

Problem 46

Textbook Question

A small car with mass $0.800$ kg travels at constant speed on the inside of a track that is a vertical circle with radius $5.00$ m (Fig. E $5.45$ ). If the normal force exerted by the track on the car when it is at the top of the track (point $B$ ) is $6.00$ N, what is the normal force on the car when it is at the bottom of the track (point $A$ )?

Verified step by step guidance

Step 1: Identify the forces acting on the car at the top of the track (point B). These include the gravitational force (mg) acting downward and the normal force (N_top) exerted by the track, also acting downward. The net force at the top is the centripetal force required to keep the car moving in a circular path.

Step 2: Write the equation for the net force at the top of the track using Newton's second law. The centripetal force is provided by the sum of the gravitational force and the normal force:

F_{net} = N_{top} + m g

. The centripetal force is also expressed as

m v^{2} / r

, where v is the speed of the car and r is the radius of the circular track.

Step 3: Solve for the speed of the car (v) using the forces at the top of the track. Rearrange the equation:

v^{2} = r (N_{top} + m g)/ m

. Substitute the given values for mass (m = 0.800 kg), radius (r = 5.00 m), normal force at the top (N_top = 6.00 N), and gravitational acceleration (g = 9.80 m/s²).

Step 4: At the bottom of the track (point A), the forces acting on the car are the gravitational force (mg) acting downward and the normal force (N_bottom) exerted by the track acting upward. The net force at the bottom is again the centripetal force required to keep the car moving in a circular path. Write the equation for the net force at the bottom:

N_{bottom} - m g = m v^{2} / r

Step 5: Solve for the normal force at the bottom (N_bottom) by rearranging the equation:

N_{bottom} = m v^{2} / r + m g

. Substitute the values for mass, radius, gravitational acceleration, and the speed of the car (v) calculated in Step 3 to find the normal force at the bottom.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Centripetal Force

Centripetal force is the net force acting on an object moving in a circular path, directed towards the center of the circle. For an object in vertical circular motion, this force is crucial for maintaining the circular trajectory. At the top of the circle, the gravitational force and the normal force both contribute to the centripetal force, while at the bottom, the normal force must overcome gravity to provide the necessary centripetal force.

Normal Force

The normal force is the force exerted by a surface to support the weight of an object resting on it, acting perpendicular to the surface. In the context of circular motion, the normal force varies depending on the position of the object in the circle. At the top of the track, the normal force is reduced due to the gravitational force acting in the same direction, while at the bottom, it must counteract gravity and provide additional force for centripetal acceleration.

Newton's Second Law

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass, expressed as F = ma. This principle is essential for analyzing forces in circular motion, as it allows us to calculate the net force required for the car to maintain its circular path. By applying this law at different points in the circular motion, we can determine the normal force at the bottom of the track.

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