Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 16 - Planar Kinematics of a Rigid Body - Section 16.3 - Rotation about a Fixed Axis - Problems - Page 331: 1

Answer

$v_A=2.6m/s$, $a_A=9.35m/s^2$

Work Step by Step

We can determine the required velocity and acceleration as follows: $v_A=\omega r$ $\implies v_A=(5t^2+2)r$ We plug in the known values to obtain: $v_A=[5(0.5)^2+2](0.8)$ This simplifies to: $v_A=2.6m/s$ We know that $a_z=ar=(10t)r=10(0.5)(0.8)=4m/s^2$ and $a_n=(5t^2+2)^2r=[5(0.5)^2+2]^2(0.8)=8.45m/s^2$ Now, $a_A=\sqrt{a_z^2+a_n^2}$ We plug in the known values to obtain: $a_A=\sqrt{(4)^2+(8.45)^2}$ This simplifies to: $a_A=9.35m/s^2$
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