I would Like step by step solutions please. I REALLY NEED STEP BY STEP A student

Question

I would Like step by step solutions please. I REALLY NEED STEP BY STEP

A student tries to avoid taking a physics test by attempting to stop the rotation of the Earth, thus preventing Friday from arriving. The student does so by taking a very large rocket and firmly attaching it to the ground at the equator, so that the engine is horizontal and the nose of the rocket points west. If the rocket can generate a force of 1.76 Times 107 N, how long will it have to run in order to bring the Earth's rotation to a halt? (Note: the mass of the Earth is 5.98 Times 10 24 kg. the radius of the Earth is 6.37 Times 106 m, and the moment of inertia of a solid sphere rotating about an axis through its center of mass is 2 / 5mr2. Also, considering the fact that the student is behaving rather irrationally, the usual expectation of physically realistic answers may not apply in this case.)

Explanation / Answer

Given that The horizontal force, F = 1.76 * 107 N,   Radius of earth, R = 6378.1 km = 6378100 m

Mass of earth, M = 5.97 * 1024 kg

Initial angular velocity of earth, w1 = (2pi / 86400) rad/s = 7.27 * 10-5 rad/s

Final angular velcoity of earth, w2 = 0

Moment of inertia of earth, I = (2/5)MR2

We have a formula for torque as T= I alpha

FR = I Alpha

FR = [(2/5)MR2][ (w2-w1)/dt ]

dt= [(2/5)MR][ (w2-w1)/F ]

By substituting all the values

dt = [(2/5)(5.97 * 1024 kg)(6378100 m)][ (0-7.27 * 10-5 rad/s)/1.76 * 107 N ]

dt = 6.29 * 1019 s Given that The horizontal force, F = 1.76 * 107 N,   Radius of earth, R = 6378.1 km = 6378100 m

Mass of earth, M = 5.97 * 1024 kg

Initial angular velocity of earth, w1 = (2pi / 86400) rad/s = 7.27 * 10-5 rad/s

Final angular velcoity of earth, w2 = 0

Moment of inertia of earth, I = (2/5)MR2

We have a formula for torque as T= I alpha

FR = I Alpha

FR = [(2/5)MR2][ (w2-w1)/dt ]

dt= [(2/5)MR][ (w2-w1)/F ]

By substituting all the values

dt = [(2/5)(5.97 * 1024 kg)(6378100 m)][ (0-7.27 * 10-5 rad/s)/1.76 * 107 N ]

dt = 6.29 * 1019 s Given that The horizontal force, F = 1.76 * 107 N,   Radius of earth, R = 6378.1 km = 6378100 m

Mass of earth, M = 5.97 * 1024 kg

Mass of earth, M = 5.97 * 1024 kg

Initial angular velocity of earth, w1 = (2pi / 86400) rad/s = 7.27 * 10-5 rad/s

Final angular velcoity of earth, w2 = 0

Moment of inertia of earth, I = (2/5)MR2

We have a formula for torque as T= I alpha

FR = I Alpha

FR = [(2/5)MR2][ (w2-w1)/dt ]

dt= [(2/5)MR][ (w2-w1)/F ]

By substituting all the values

dt = [(2/5)(5.97 * 1024 kg)(6378100 m)][ (0-7.27 * 10-5 rad/s)/1.76 * 107 N ]

dt = 6.29 * 1019 s dt = [(2/5)(5.97 * 1024 kg)(6378100 m)][ (0-7.27 * 10-5 rad/s)/1.76 * 107 N ]

dt = 6.29 * 1019 s dt = 6.29 * 1019 s

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