CP CALC? A disk of radius 25.0 cm is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (?Fig. P9.59?). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation a(t) = At, where t is in seconds and A is a constant. The cylinder starts from rest, and at the end of the third second, the ball’s acceleration is 1.80 m/s2. (a) Find A. (b) Express the angular acceleration of the disk as a function of time. (c) How much time after the disk has begun to turn does it reach an angular speed of 15.0 rad/s? (d) Through what angle has the disk turned just as it reaches 15.0 rad/s? (?Hint:? See Section 2.6.)

Solution 65P Introduction First we have to find out the constant A, from this we can find out the linear acceleration as a function of time. From this functional form we can calculate the rotational acceleration as a function of time. Then we have to calculate the angular velocity and displacement after given time. Step 1 The linear acceleration is given by a(t) = At 2 Now at t = 3 s we have a = 1.80 m/s , hence we have 2 A = t = 13.00 ss = 0.600 m/s 3 Hence the constant A is given by 0.600 m/s 3. Step 2 Now the linear acceleration of the edge of the disc will be equal to the linear acceleration of the ball, which is given by a(t) = (0.600 m/s )t Now the angular acceleration is given by 3 = a(t)= (0.600 m/s )= (2.40 rad.s )t3 r (25.0×102m)