Physics 1 Problem Solutions 76

# Physics 1 Problem Solutions 76 - 72 Chapt 6 Newton’s Laws...

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Unformatted text preview: 72 Chapt. 6 Newton’s Laws and Dynamics II VELOCITY (radians per second) of the Earth at the equator. What values does he get? HINT: Waldo does remember about the circumference of a circle. b) Now Waldo calculates the magnitude of centripetal acceleration for a body on the equator. What does he find? c) The sagacious swine now reflects on inertial frames. For most purposes the Earth’s surface is a pretty good inertial frame. But the Earth is rotating, and so it’s not exactly an inertial frame. The fixed stars (as they are called for historical reasons) define a more inertial frame. Say we take—Waldo says to himself— vectora in Newton’s 2nd law vector F net = mvectora and decompose it (as nothing forbids us to do) into acceleration relative to the rotating frame of equatorial Ecuador (i.e., vectora ′ ) and centripetal acceleration ( vectora c = − a c ˆ r , where ˆ r is a radially outward pointing unit vector) which is the acceleration of the rotating frame. We obtain vectora...
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