Answer
$2.6\times 10^{-4}Kg.\frac{m^2}{s}$
Work Step by Step
We know that
$L=I\omega$
$\implies L=(MR^2)(\frac{v}{R})$
$\implies L=MRv$
We plug in the known values to obtain:
$L=(0.0050)(0.095)(0.55)$
$L=2.6\times 10^{-4}Kg.\frac{m^2}{s}$
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 11 - Rotational Dynamics and Static Equilibrium - Problems and Conceptual Exercises - Page 370: 61
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