thus far in the course we have ignored that on the earths surface an object must have a slightly unbalanced force to rotate with the earth, pretending that the normal force precisely balances the objects weight. A student with a mass of 65.0kg stands at the equator. The radius of the earth is 6.38*10^6m and of course it rotates once per day.
A) what is the magnitude of the centripetal force (in newtons) required to keep the student on the earths surface?
B) Since the centripetal force is unbalanced, what is the true magnitude of the normal force on the student? (the student weighs 637 N, and with EF=ma you should find the normal force is actually less than 637 N.)
This question begs for correction, as it is based on faulty reasoning.
The centripetal force is about the center of rotation, not the center of the earth. When a mass is placed in orbit, take the moon for instance, both the moon and the Earth rotate about the center of rotation, not Earths center. Both have centripetal forces about the center of rotation, and yes, they are on opposite sides of the center of rotation, and guess what...they are equal, and balance. You can do the math and prove it yourself, and that case is identical to this, the man is is not inorbit, however, his presence changes the system center of mass, and the center of rotation. He rotates about the center of rotation, not Earths center (albeit the difference is insignificantly small), and the Earth rotates about that center as well, and ghe forces do equal, and they are in fact balanced. Newtons third law (forces are in pairs) holds.
To solve this problem, we need to consider the forces acting on the student and use the concept of centripetal force. Let's break it down step by step:
A) Magnitude of the centripetal force:
The centripetal force is the force that keeps an object moving in a circular path. In this case, it is the force that keeps the student on the Earth's surface while it is rotating.
We can find the centripetal force using the formula: Fc = m * ω^2 * r, where Fc is the centripetal force, m is the mass of the student, ω is the angular velocity, and r is the radius.
Given:
Mass of the student (m) = 65.0 kg
Angular velocity (ω) = 2π radians/day (since the Earth completes one rotation per day)
Radius (r) = 6.38 * 10^6 m
We need to convert the angular velocity from radians per day to radians per second, since the SI unit for time is second. There are 24 hours in a day, and 60 minutes in an hour, and 60 seconds in a minute, so there are 24 * 60 * 60 = 86,400 seconds in a day.
Angular velocity (ω) = 2π radians/day * (1 day / 86,400 seconds) = 2π / 86,400 radians/second
Now, plug in the values into the formula:
Fc = 65.0 kg * (2π / 86,400 radians/second)^2 * 6.38 * 10^6 m
Calculate the centripetal force using a calculator:
Fc ≈ 201.4 Newtons (rounded to one decimal place)
Therefore, the magnitude of the centripetal force required to keep the student on the Earth's surface is approximately 201.4 Newtons.
B) True magnitude of the normal force:
Since the student is not falling through the Earth's surface, there must be an equal and opposite force to counterbalance the weight of the student. Normally, we would think that the normal force is equal to the weight of the student, which is 637 N. However, in reality, due to the unbalanced centripetal force required to keep the student on the rotating Earth, the magnitude of the normal force is actually less than 637 N.
To find the true magnitude of the normal force, we can use Newton's second law of motion, which states that the net force (F_net) acting on an object is equal to the mass (m) times the acceleration (a): F_net = m * a.
In this case, the net force is the difference between the weight of the student and the centripetal force: F_net = F_weight - Fc.
Given:
Weight of the student (F_weight) = 637 N (given in the question)
Centripetal force (Fc) ≈ 201.4 N (calculated in part A)
F_net = 637 N - 201.4 N
F_net ≈ 435.6 N
Since the net force (F_net) is equal to the mass (m) times the acceleration due to gravity (g), we can rewrite the equation as:
m * g = 435.6 N
Rearranging the equation to find the true magnitude of the normal force (F_normal):
F_normal = m * g
Given:
mass of the student (m) = 65.0 kg
acceleration due to gravity (g) = 9.8 m/s^2
Plug in the values:
F_normal = 65.0 kg * 9.8 m/s^2
Calculate the true magnitude of the normal force using a calculator:
F_normal ≈ 637 N (rounded to one decimal place)
Therefore, the true magnitude of the normal force on the student is approximately 637 N, which is equal to the weight of the student.