CALC? A uniform disk with radius R = 0.400 m and mass 30.0 kg rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to ?(t) = (1.10 rad/s)t + (6.30 rad/s2)t2. What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.100 rev?
Solution 62P Step 1: 2 2 Provided, (t) = 1.10 rad/s t + (6.30 rad/s ) t Take the first derivative of with respect to time and it will give you the angular velocity, ’ (t) = 1.10 rad/s + (12.60 rad/s ) t 2 Take the second derivative of with respect to time and it will give you the angular acceleration, ” (t) = 0 + (12.60 rad/s ) 2 2 Therefore, tangential acceleration, a = 12.60 rad/s