A wheel rotating about a fixed axis with a constant angular acceleration of 2.0 rad/s² starts from rest at t = 0.The wheel has a diameter of 20 cm.What is the magnitude of the total linear acceleration of a point on the outer edge of the wheel at t = 0.60 s? A)0.24 m/s² B)0.50 m/s² C)0.14 m/s² D)0.34 m/s² E)0.20 m/s²

Question

Quizplus · Accepted Answer

Explanation:The total linear acceleration of a point on the outer edge of the wheel is the vector sum of the tangential acceleration and the centripetal acceleration. The tangential acceleration is given by:$$a_t = rα$$where r is the radius of the wheel and α is the angular acceleration. The centripetal acceleration is given by:$$a_c = v^2/r$$where v is the tangential velocity. At t = 0, the wheel is at rest, so v = 0. Therefore, the total linear acceleration is equal to the tangential acceleration:$$a = a_t = rα = (0.10 	ext{ m})(2.0 	ext{ rad/s}^2) = 0.20 	ext{ m/s}^2$$

A Wheel Rotating About a Fixed Axis with a Constant