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Chapter 5: Problem 46

\(\mathrm{A}\) small car with mass \(0.800 \mathrm{~kg}\) travels at constant speed on the inside of a track that is a vertical circle with radius \(5.00 \mathrm{~m}\) (Fig. E5.45). If the normal force exerted by the track on the car when it is at the top of the track (point \(B\) ) is \(6.00 \mathrm{~N}\), what is the normal force on the car when it is at the bottom of the track (point \(A\) )?

Short Answer

Expert verified
The normal force on the car when it is at the bottom of the track is \(21.68N\).

Step by step solution

01

Identify the forces at the top point B

At the top of the circle (point \(B\)), the forces acting on the car are the gravitational force and the normal force. These forces are acting in the same downwards direction. Gravity (weight of the car) is calculated by \[F = m \cdot g\] where \(m = 0.8kg\) and \(g = 9.8m/s^2\) which equals \(7.84N\]. The normal force at point B is provided in the problem as \(6N\). Therefore, the resultant force at the top \(F_{top}\) is equal to gravity plus the normal force, hence \(F_{top} = 6N + 7.84N = 13.86N\].
02

Calculate the Centripetal Force

[Centripetal Force](https://en.wikipedia.org/wiki/Centripetal_force) is the net force that keeps an object moving in a circular path. It can be calculated using the formula \[F_{c} = m \cdot g\] since the car is moving at a constant speed that implies the net force acting on the car when it's at the top is equal to the centripetal force. Hence, at the top of the circle, \(F_{c_{top}} = F_{top} = 13.84N\].
03

Identify the forces at the bottom point A

At the bottom of the circle (point \(A\)), the forces acting on the car are again the gravitational force and the normal force. This time, however, they are acting in opposite directions. The gravitational force is in the downward direction and the normal force is in the upward direction, preventing the car from falling through the track.
04

Calculate the Normal Force at point A

Given that the speed and therefore the centripetal force \(F_c\) does not change around the loop, at the bottom of the path, the net force or centripetal force equals the normal force subtracted by the weight of the car. Thus, we solve for the normal force at the bottom \(F_{A}\) using the equation \(F_{c_{bottom}} = F_{A} - gravity\) and we find that \(F_{A} = F_{c_{bottom}} + gravity = 13.84N + 7.84N = 21.68N\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Force
Centripetal force is crucial in understanding circular motion. It is the force required to keep an object moving in a circular path, always directed towards the center of the circle. For an object moving in a circle, like our car on the track, this force is what keeps it from flying off. The formula for centripetal force is given by \( F_c = \frac{mv^2}{r} \), where \( m \) is mass, \( v \) is velocity, and \( r \) is the radius of the circle.
  • It acts perpendicular to the object's velocity.
  • Its magnitude is dependent on mass, speed, and radius.
In the problem, we determine that the centripetal force at the top of the track is the sum of the gravitational and normal forces acting downwards. This is a key realization for solving problems involving vertical circular motion.
Normal Force
The normal force is a contact force exerted perpendicular to the surface of contact. In the context of our circular track, it is the force exerted by the track on the car.

  • At the top of the circle, the normal force acts in the same direction as gravity.
  • At the bottom of the circle, the normal force opposes gravity, pushing upwards.
Understanding the direction relative to gravitational force is key. Picture a roller coaster. When reaching the top, both gravity and track push the coaster down. At the bottom, the track pushes the coaster up against gravity. Calculating normal force requires knowing the other forces and how they sum with the centripetal force needed for motion.
Gravitational Force
Gravitational force, or weight, is the force exerted by gravity on an object. It acts downwards and is calculated using the formula \( F_g = m \cdot g \), where \( g = 9.8 \text{ m/s}^2 \).
  • Consistently acts towards the center of the earth.
  • Magnitude depends on object's mass.
In circular motion problems, distinguishing gravitational from other forces is crucial. For instance, when the car is at the top of the track, gravity and normal force add together to create the required centripetal force. Thus, understanding its role helps solve force-related equations in dynamics.
Vertical Circular Motion
Vertical circular motion involves an object moving along a vertical circular path, facing varying forces at different points.

  • At the top, forces like gravity aid the circular motion.
  • At the bottom, opposing forces such as normal force help counteract gravity.
This kind of motion is complicated by the gravity acting at all points, creating varying normal forces. For a car speeding around a track, its speed must be sufficient to maintain contact with the track, relying on gravity and velocity to provide the necessary centripetal force at various positions. Visualizing the movement and relating it to force dynamics is central to mastering these problems.

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Most popular questions from this chapter

\(\mathrm{CP}\) A \(5.00 \mathrm{~kg}\) box sits at rest at the bottom of a ramp that is \(8.00 \mathrm{~m}\) long and is inclined at \(30.0^{\circ}\) above the horizontal. The coefficient of kinetic friction is \(\mu_{\mathrm{k}}=0.40,\) and the coefficient of static friction is \(\mu_{\mathrm{s}}=0.43 .\) What constant force \(F,\) applied parallel to the surface of the ramp, is required to push the box to the top of the ramp in a time of \(6.00 \mathrm{~s} ?\)

Rotating Space Stations. One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates "artificial gravity" at the outside rim of the station. (a) If the diameter of the space station is \(800 \mathrm{~m}\), how many revolutions per minute are needed for the "artificial gravity" acceleration to be \(9.80 \mathrm{~m} / \mathrm{s}^{2} ?\) (b) If the space station is a waiting area for travelers going to Mars, it might be desirable to simulate the acceleration due to gravity on the Martian surface \(\left(3.70 \mathrm{~m} / \mathrm{s}^{2}\right) .\) How many revolutions per minute are needed in this case?

Genesis Crash. On September \(8,2004,\) the Genesis spacecraft crashed in the Utah desert because its parachute did not open. The \(210 \mathrm{~kg}\) capsule hit the ground at \(311 \mathrm{~km} / \mathrm{h}\) and penetrated the soil to a depth of \(81.0 \mathrm{~cm}\). (a) What was its acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(g\) 's) assumed to be constant, during the crash? (b) What force did the ground exert on the capsule during the crash? Express the force in newtons and as a multiple of the capsule's weight. (c) How long did this force last?

Two blocks, with masses \(4.00 \mathrm{~kg}\) and \(8.00 \mathrm{~kg},\) are connected by a string and slide down a \(30.0^{\circ}\) inclined plane (Fig. \(\mathbf{P 5 . 9 6}\) ). The coefficient of kinetic friction between the \(4.00 \mathrm{~kg}\) block and the plane is \(0.25 ;\) that between the \(8.00 \mathrm{~kg}\) block and the plane is \(0.35 .\) Calculate (a) the acceleration of each block and (b) the tension in the string. (c) What happens if the positions of the blocks are reversed, so that the \(4.00 \mathrm{~kg}\) block is uphill from the \(8.00 \mathrm{~kg}\) block?

A \(75.0 \mathrm{~kg}\) wrecking ball hangs from a uniform, heavy-duty chain of mass \(26.0 \mathrm{~kg}\). (a) Find the maximum and minimum tensions in the chain. (b) What is the tension at a point three-fourths of the way up from the bottom of the chain?
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