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 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 total acceleration of a point on a rotating object is the vector sum of the centripetal acceleration and the tangential acceleration. The centripetal acceleration is given by:$$a_c = \frac{v^2}{r}$$where v is the tangential velocity and r is the distance from the point to the axis of rotation. The tangential acceleration is given by:$$a_t = r\alpha$$where $\alpha$ is the angular acceleration.In this problem, the CD has a diameter of 12.0 cm, so the radius is 6.00 cm. The CD starts from rest, so the initial tangential velocity is 0. The angular acceleration is 0.600 rad/sec^2. After 1.30 sec, the angular velocity is:$$\omega = \alpha t = (0.600 	ext{ rad/sec}^2)(1.30 	ext{ s}) = 0.780 	ext{ rad/s}$$The tangential velocity is:$$v = r\omega = (0.0600 	ext{ m})(0.780 	ext{ rad/s}) = 0.0468 	ext{ m/s}$$The centripetal acceleration is:$$a_c = \frac{v^2}{r} = \frac{(0.0468 	ext{ m/s})^2}{0.0600 	ext{ m}} = 0.0392 	ext{ m/s}^2$$The tangential acceleration is:$$a_t = r\alpha = (0.0600 	ext{ m})(0.600 	ext{ rad/sec}^2) = 0.0360 	ext{ m/s}^2$$The total acceleration is:$$a = \sqrt{a_c^2 + a_t^2} = \sqrt{(0.0392 	ext{ m/s}^2)^2 + (0.0360 	ext{ m/s}^2)^2} = 0.0532 	ext{ m/s}^2$$Therefore, the correct answer is A.

A CD Has a Diameter of 12