We note that the object turns downwards which implies

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We note that the object turns downwards which implies that the the radial acceleration points downwards. So the sum of the forces in the ˆ y direction is N W . By Newton’s 2nd law, the sum of the forces in the ˆ y direction is the radial force and is proportional to the radial acceleration. The radial acceleration points downwards, so the radial force points downwards and thus W N . We know that kinetic friction always opposes the velocity, so it must point in the negative ˆ x direction at the point shown. Again by Newton’s 2nd law, the sum of the forces in the ˆ x direction F ma x , where a x dv x dt . Since the only force in the ˆ x direction is friction, and v points totally in the x direction, we have F Net ma x f m dv x dt . These two conditions give the solution a. 2. Due to tidal forces generated by the gravitational pull of the moon on the oceans, the length of the day on Earth is slowly increasing over time. Let the difference between the apparent gravitational acceleration at the North pole and at the equator be Δ g g pole g equator . If over a million years the length of the day increases by 1%, (a) Δ g decreases by 2%. 52 (b) Δ g decreases by 1%. (c) Δ g decreases by 0.5%. (d) Δ g increases by 2%. (e) Δ g increases by 1%. 3
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We typically think of the acceleration due to gravity as the total down- wards acceleration experienced on the surface of the planet. We learned earlier this year that bodies undergoing uniform circular motion experience a centripetal force. This week we learned that in the reference frame of the body, objects don’t experience the centripetal force and instead expe- rience the fictitious, or non-inertial, force with the same magnitude as the centripetal force, but pointing in the opposite direction. Applying this to objects on the surface of the earth we find that the apparent force of gravity is: W 2 R, where m is the mass of the object we are analyzing, ω is the rotational velocity of the earth, and R is the distance between the axis of rotation to the point on the surface of the earth. At the equator, the distance between the axis of rotation and the surface is equal to the radius of the earth. If we were to stand at the pole, then we are standing on the axis of rotation, and thus R 0, so F pole W 2 R W 2 0 W. Then we have that F pole W , and F equator W 2 R , where R is the radius of the earth. For instance, consider a tennis ball of mass m on the surface of the earth. The difference between the apparent force of gravity at the pole and the equator is: F pole F equator Δ F W W 2 R m Δ a mg mg 2 R m Δ a Diving through by m , we find that 4
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g g ω 2 R ω 2 R Δ a. Which shows that the mass of the object is irrelevant, so we can consider the general case of any object on the earth and find the apparent gravitational acceleration.
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