By Kleppner D., Kolenkow R.

ISBN-10: 0521198119

ISBN-13: 9780521198110

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Additional info for An Introduction to Mechanics

Example text

8 we discussed the motion given by r = r (cos ωt ˆi + sin ωt ˆj). The velocity is v = r ω(− sin ωt ˆi + cos ωt ˆj). Because r · v = r2 ω(− cos ωt sin ωt + sin ωt cos ωt) =0 v r we see that dr/dt is perpendicular to r. We conclude that the magnitude of r is constant. Consequently, the only possible change in r is a change in its direction, which is to say that r must rotate and the trajectory is a circle, This is precisely the case: r rotates about the origin. We showed earlier that a = −ω2 r. Since r · v = 0, it follows that a · v = −ω2 r · v = 0 and a = dv/dt is perpendicular to v.

The second term, however, involves a new concept—taking the time derivative of a base vector. So, let us investigate how to do this, ˆ both for rˆ (θ) and for θ(θ). 2 dˆr/dt and dθ/dt in Polar Coordinates ˆ We will need Our goal here is to calculate the time derivatives of rˆ and θ. these results to express velocity v and acceleration a in polar coordinates. Using Newton’s notation for time derivatives can help make equations easier to read. For example, dθ ˙ =θ dt d2 θ ¨ = θ. dt2 Our starting point is Eq.

8 we discussed the motion given by r = r (cos ωt ˆi + sin ωt ˆj). The velocity is v = r ω(− sin ωt ˆi + cos ωt ˆj). Because r · v = r2 ω(− cos ωt sin ωt + sin ωt cos ωt) =0 v r we see that dr/dt is perpendicular to r. We conclude that the magnitude of r is constant. Consequently, the only possible change in r is a change in its direction, which is to say that r must rotate and the trajectory is a circle, This is precisely the case: r rotates about the origin. We showed earlier that a = −ω2 r. Since r · v = 0, it follows that a · v = −ω2 r · v = 0 and a = dv/dt is perpendicular to v.