Physics 1 Problem Solutions 76

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online