A CD has a diameter of 12.0 cm. If the CD starts from rest and has a constant angular acceleration of 0.600 rad/sec², then the magnitude of the (total) acceleration of a point 3.00 cm from the center of the CD after 1.30 sec is A) 0.0256 m/s². B) 0.0306 m/s². C) 0.0415 m/s². D) 0.0472 m/s². E) 0.0502 m/s².

Question

Quizplus · Accepted Answer

Explanation:The tangential acceleration is given by:$$a_t = r \alpha$$where:r is the distance from the center of rotationα is the angular accelerationThe centripetal acceleration is given by:$$a_c = \frac{v^2}{r}$$where:v is the tangential velocityThe total acceleration is given by:$$a = \sqrt{a_t^2 + a_c^2}$$Substituting the given values, we get:$$a_t = (0.0300 \ m)(0.600 \ rad/s^2) = 0.0180 \ m/s^2$$$$a_c = \frac{(0.0180 \ m/s)^2}{0.0300 \ m} = 0.0108 \ m/s^2$$$$a = \sqrt{(0.0180 \ m/s)^2 + (0.0108 \ m/s)^2} = 0.0216 \ m/s^2$$Therefore, the magnitude of the (total) acceleration of a point 3.00 cm from the center of the CD after 1.30 sec is 0.0216 m/s².

A CD Has a Diameter of 12