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A student said, "When the Moon goes around the Earth, there is an inward force due to the Moon and an outward force due to centrifugal force, so the net force on the Moon is zero." Give two or more physics reasons why this is wrong.

Short Answer

Expert verified

The net force is non-zero.

Step by step solution

01

Definition of centrifugal force

In a circular motion, centrifugal force works along the radius and is directed away from the circle's centre. When measuring in an inertial frame of reference, the force does not exist. It only becomes relevant when we switch from a ground/inertial to a spinning reference frame.

02

Finding the net force

The net force on an object can be expressed as the sum of two parts and it is equal to the rate of change of the momentum.

The two parts that we are taken about are the parallel rate of change of the momentum dpdtand the perpendicular rate of change of the momentum dpdt.

Write the net force Fneton the object.

Fnet=dpdt=dpdt-dpdt

03

Analysing the momentum

The moon experiences an inward gravitational attraction as it revolves around the Earth. Although there is an outward centrifugal force, it is not a true force. on equating them, the resulting will be a non-zero net force on the moon. As a result, the stated assertion is incorrect.

The object is moving in a curve path, and its momentum is constantly changing. Even if the object moves with constant speed, a change of direction indicates that dpdtis non zero. Hence the net force is non zero.

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Most popular questions from this chapter

An object moving at a constant speed of 23m/sis making a turn with a radius of curvature of 4m(this is the radius of the kissing circle). The object's momentum has a magnitude of . What is the magnitude of the rate of change of the momentum? What is the magnitude of the net force?

A helicopter flies to the right (in the +xdirection) at a constant speed of 12m/s, parallel to the surface of the ocean. A 900 kgpackage of supplies is suspended below the helicopter by a cable as shown in Figure the package is also traveling to the right in a straight line, at a constant speed of 12 m/s. The pilot is concerned about whether or not the cable, whose breaking strength is listed at 9300 N , is strong enough to support this package under these circumstances.

(a) Choose the package as the system, and draw a free-body diagram.

(b) What is the magnitude of the tension in the cable supporting the package?

(c) Write the force exerted on the package by the cable as a vector.

(d) What is the magnitude of the force exerted by the air on the package?

(e) Write the force on the package by the air as a vector.

(f) Is the cable in danger of breaking?

Use a circular pendulum to determine . You can increase the accuracy of the time it takes to go around once by timingN revolutions and then dividing by N. This minimizes errors contributed by inaccuracies in starting and stopping the clock. It is wise to start counting from zero (0,1,2,3,4,5)rather than starting from (0,1,2,3,4,5)represents only four revolutions, not five). It also improves accuracy if you start and stop timing at a well-defined event, such as when the mass crosses in front of an easily visible mark. This was the method used by Newton to get an accurate value of g. Newton was not only a brilliant theorist but also an excellent experimentalist. For a circular pendulum, he built a large triangular wooden frame mounted on a vertical shaft, and he pushed this around and around while making sure that the string of the circular pendulum stayed parallel to the slanting side of the triangle.

Tarzan swings from a vine. When he is at the bottom of his swing, as shown in Figure 5.63, which is larger in magnitude: the force by the Earth on Tarzan, the force by the vine (a tension force) on Tarzan, or neither (same magnitude)? Explain how you know this.

Question: A student said, "When the Moon goes around the Earth, there is an inward force due to the Moon and an outward force due to centrifugal force, so the net force on the Moon is zero." Give two or more physics reasons why this is wrong.

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