Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 11 - Rolling, Torque, and Angular Momentum - Problems - Page 322: 27a

Answer

The angular momentum about the origin is $~~0$

Work Step by Step

We can write the general form of a cross-product: $a\times b = (a_yb_z-b_ya_z)~\hat{i}+(a_zb_x-b_za_x)~\hat{j}+(a_xb_y-b_xa_y)~\hat{k}$ We can express the particle's position in unit-vector notation: $r = (2.0~m)~\hat{i}+ (-2.0~m)~\hat{k}$ We can express the velocity in unit-vector notation: $v = (-5.0~m/s)~\hat{i}+(5.0~m/s)~\hat{k}$ We can find the angular momentum about the origin: $L = m~(r \times v)$ $L = (0.25~kg)~[(0)(5.0~m/s)-(0)(-2.0~m)]~\hat{i}+[(-2.0~m)(-5.0~m/s)-(5.0~m/s)(2.0~m)]~\hat{j}+[(2.0~m)(0)-(-5.0~m/s)(0)]~\hat{k}$ $L = (0.25~kg)~[(0)~\hat{i}+(0)~\hat{j}+(0)~\hat{k}]$ $L = 0$ The angular momentum about the origin is $~~0$
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