## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 65

#### Answer

(a) The magnitude of the acceleration is $17600~m/s^2$ (b) The magnitude of the acceleration is $4430~m/s^2$

#### Work Step by Step

(a) We can convert 4000 rpm to units of rad/s $\omega = (4000~rpm)(\frac{1~min}{60~s})(\frac{2\pi~rad}{1~rev})$ $\omega = 419~rad/s$ We can find the centripetal acceleration at the end of a test tube. $a_c = \omega^2~r$ $a_c = (419~rad/s)^2(0.10~m)$ $a_c = 17600~m/s^2$ The magnitude of the acceleration is $17600~m/s^2$ (b) We can find the speed just before hitting the floor. $v^2= v_0^2+2ay$ $v = \sqrt{2ay} = \sqrt{(2)(9.8~m/s^2)(1.0~m)}$ $v = 4.43~m/s$ We can find the rate of deceleration after striking the floor. We can let $v_0 = 4.43~m/s$ for this part of the question. $a = \frac{v-v_0}{t}$ $a = \frac{0-4.43~m/s}{1.0\times 10^{-3}~s}$ $a = -4430~m/s^2$ The magnitude of the acceleration is $4430~m/s^2$

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