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 334: 25


$v_P=2.42~ft/s$ $a_P=34.4~ft/s^2$

Work Step by Step

The required velocity and acceleration can be determined as follows: As $\alpha=400 t^3$ $\implies 400t^3=\frac{d\omega}{dt}$ $\implies d\omega=400t^3dt$ $\implies \int_0^{\omega_A} d\omega\implies \int_0^t 400 t^3 dt$ $\implies \omega|_0^{\omega_A}=(\frac{400t^4}{4})|_0^t$ $\implies \omega_A=100 t^4$ Similarly, $\omega_B=\frac{\omega_A r_A}{r_B}=\frac{(31.6406)(0.5)}{1.2}=13.1836rad/s$ and $\alpha_B=\frac{\alpha_A r_A}{r_B}=\frac{(168.75)(0.5)}{1.2}=70.3~rad/s^2$ Now the velocity at point B is given as $v_P=\omega_B r_P$ $\implies v_P=13.1836\times 2.2=29in/s =29/12=2.4169ft/s$ We know that $a_{Pt}=\alpha_B r_P$ $\implies a_{Pt}=70.3125\times \frac{2.2}{12}=12.8906ft/s^2$ and $a_{Pn}=\omega^2_B r$ $\implies a_{Pn}=(13.1836)^2(\frac{2.2}{12})=31.8646ft/s^2$ Now $a_P=\sqrt{(a_{Pt})^2+(a_{Pn})^2}$ We plug in the known values to obtain: $a_P=\sqrt{(12.8906)^+(31.8646)^2}=34.4~ft/s^2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.