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A small body of mass m = 0.30 kg starts sliding down from the top of a smooth sphere of radius R = 1.00 m. The sphere rotates with a constant angular velocity ω = 6.0 rad/s about a vertical axis passing through its centre. Find the centrifugal force of inertia and the Coriolis force at the moment when the body breaks off the surface of the sphere in the reference frame fixed to the sphere.?
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A small body of mass m = 0.30 kg starts sliding down from the top of a...
Centrifugal Force of Inertia:
The centrifugal force of inertia is the apparent force experienced by an object moving in a rotating reference frame. It is given by the equation:
Fci = m * R * ω^2
where:
- Fci is the centrifugal force of inertia
- m is the mass of the object
- R is the radius of the sphere
- ω is the angular velocity of the sphere
In this case, the mass of the body is given as m = 0.30 kg, the radius of the sphere is R = 1.00 m, and the angular velocity is ω = 6.0 rad/s. Substituting these values into the equation, we can calculate the centrifugal force of inertia:
Fci = 0.30 kg * 1.00 m * (6.0 rad/s)^2
Fci = 10.8 N
Therefore, the centrifugal force of inertia at the moment when the body breaks off the surface of the sphere is 10.8 N.
Coriolis Force:
The Coriolis force is another apparent force experienced by an object moving in a rotating reference frame. It acts perpendicular to the velocity of the object and is given by the equation:
Fc = 2 * m * v * ω
where:
- Fc is the Coriolis force
- m is the mass of the object
- v is the velocity of the object
- ω is the angular velocity of the sphere
At the moment when the body breaks off the surface of the sphere, its velocity is tangential to the sphere's surface. The tangential velocity can be calculated using the equation:
v = R * ω
Substituting the given values, we find:
v = 1.00 m * 6.0 rad/s
v = 6.0 m/s
Now, substituting the values of mass (m = 0.30 kg), velocity (v = 6.0 m/s), and angular velocity (ω = 6.0 rad/s) into the equation for the Coriolis force, we can calculate it as follows:
Fc = 2 * 0.30 kg * 6.0 m/s * 6.0 rad/s
Fc = 21.6 N
Therefore, the Coriolis force at the moment when the body breaks off the surface of the sphere is 21.6 N.
The centrifugal force of inertia is the apparent force experienced by an object moving in a rotating reference frame. It is given by the equation:
Fci = m * R * ω^2
where:
- Fci is the centrifugal force of inertia
- m is the mass of the object
- R is the radius of the sphere
- ω is the angular velocity of the sphere
In this case, the mass of the body is given as m = 0.30 kg, the radius of the sphere is R = 1.00 m, and the angular velocity is ω = 6.0 rad/s. Substituting these values into the equation, we can calculate the centrifugal force of inertia:
Fci = 0.30 kg * 1.00 m * (6.0 rad/s)^2
Fci = 10.8 N
Therefore, the centrifugal force of inertia at the moment when the body breaks off the surface of the sphere is 10.8 N.
Coriolis Force:
The Coriolis force is another apparent force experienced by an object moving in a rotating reference frame. It acts perpendicular to the velocity of the object and is given by the equation:
Fc = 2 * m * v * ω
where:
- Fc is the Coriolis force
- m is the mass of the object
- v is the velocity of the object
- ω is the angular velocity of the sphere
At the moment when the body breaks off the surface of the sphere, its velocity is tangential to the sphere's surface. The tangential velocity can be calculated using the equation:
v = R * ω
Substituting the given values, we find:
v = 1.00 m * 6.0 rad/s
v = 6.0 m/s
Now, substituting the values of mass (m = 0.30 kg), velocity (v = 6.0 m/s), and angular velocity (ω = 6.0 rad/s) into the equation for the Coriolis force, we can calculate it as follows:
Fc = 2 * 0.30 kg * 6.0 m/s * 6.0 rad/s
Fc = 21.6 N
Therefore, the Coriolis force at the moment when the body breaks off the surface of the sphere is 21.6 N.
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A small body of mass m = 0.30 kg starts sliding down from the top of a smooth sphere of radius R = 1.00 m. The sphere rotates with a constant angular velocity ω = 6.0 rad/s about a vertical axis passing through its centre. Find the centrifugal force of inertia and the Coriolis force at the moment when the body breaks off the surface of the sphere in the reference frame fixed to the sphere.?
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A small body of mass m = 0.30 kg starts sliding down from the top of a smooth sphere of radius R = 1.00 m. The sphere rotates with a constant angular velocity ω = 6.0 rad/s about a vertical axis passing through its centre. Find the centrifugal force of inertia and the Coriolis force at the moment when the body breaks off the surface of the sphere in the reference frame fixed to the sphere.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A small body of mass m = 0.30 kg starts sliding down from the top of a smooth sphere of radius R = 1.00 m. The sphere rotates with a constant angular velocity ω = 6.0 rad/s about a vertical axis passing through its centre. Find the centrifugal force of inertia and the Coriolis force at the moment when the body breaks off the surface of the sphere in the reference frame fixed to the sphere.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A small body of mass m = 0.30 kg starts sliding down from the top of a smooth sphere of radius R = 1.00 m. The sphere rotates with a constant angular velocity ω = 6.0 rad/s about a vertical axis passing through its centre. Find the centrifugal force of inertia and the Coriolis force at the moment when the body breaks off the surface of the sphere in the reference frame fixed to the sphere.?.
A small body of mass m = 0.30 kg starts sliding down from the top of a smooth sphere of radius R = 1.00 m. The sphere rotates with a constant angular velocity ω = 6.0 rad/s about a vertical axis passing through its centre. Find the centrifugal force of inertia and the Coriolis force at the moment when the body breaks off the surface of the sphere in the reference frame fixed to the sphere.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A small body of mass m = 0.30 kg starts sliding down from the top of a smooth sphere of radius R = 1.00 m. The sphere rotates with a constant angular velocity ω = 6.0 rad/s about a vertical axis passing through its centre. Find the centrifugal force of inertia and the Coriolis force at the moment when the body breaks off the surface of the sphere in the reference frame fixed to the sphere.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A small body of mass m = 0.30 kg starts sliding down from the top of a smooth sphere of radius R = 1.00 m. The sphere rotates with a constant angular velocity ω = 6.0 rad/s about a vertical axis passing through its centre. Find the centrifugal force of inertia and the Coriolis force at the moment when the body breaks off the surface of the sphere in the reference frame fixed to the sphere.?.
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