Solution Found!
A wheel of diameter 40.0 cm starts from rest and rotates
Chapter 11, Problem 21E(choose chapter or problem)
A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of \(3.00 \mathrm{rad} / \mathrm{s}^{2}\). At the instant the wheel has computed its second revolution, compute the radial acceleration of a point on the rim in two ways: (a) using the relationship \(a_{\mathrm{rad}}=\omega^{2} r\) and (b) from the relationship \(a_{\mathrm{rad}}=v^{2} / r\).
Questions & Answers
QUESTION:
A wheel of diameter 40.0 cm starts from rest and rotates with a constant angular acceleration of \(3.00 \mathrm{rad} / \mathrm{s}^{2}\). At the instant the wheel has computed its second revolution, compute the radial acceleration of a point on the rim in two ways: (a) using the relationship \(a_{\mathrm{rad}}=\omega^{2} r\) and (b) from the relationship \(a_{\mathrm{rad}}=v^{2} / r\).
ANSWER:Solution 21E Step 1 of 2: To calculate the the angular velocity at the end of two revolutions a) z + oz ( z ) 0 = 2 ( ) z z 0 = (3 rad/s )(4 rad) = 8.68 rad/s v2 arad = r and we know that v = r 2 so a = (r) rad r 2 = r = (0.2m)(8.68 rad/s) 2 = 15.1 m/s 2