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 Step 2: Angular displacement, = 2 × number of revolutions = 2 × 0.100 rev = 0.628 rad From the equation, ’ (t) = 1.10 rad/s + (12.60 rad/s ) t 2 We can derive the initial velocity from this relation. It is, 1.10 rad/s So, the actual equation is, ’ (t) = ’ (t)+ a 0 ’ (t) = 1.10 rad/s 0 2 2 We can use the relation, (t) = 1.10 rad/s t + (6.30 rad/s ) t since, we know the (t). So, 0.628 rad = 1.10 rad/s t + (6.30 rad/s ) t 2 2 Or, (6.30 rad/s ) t + 1.10 rad/s t - 0.628 rad = 0 This is quadratic equation and while solving it, this will provide the time. 1.10 1.10 +4×6.30×0.628 t = 2×6.30 t = 1.10 1.21 +15.826 12.60 t = 1.10 ±4.13 12.60 1.10 + 4.13 Time has only positive values, therefore, t = 12.60 t = 0.24 s Angular velocity at 0.24 s will be, ’ (t) = 1.10 rad/s + (12.60 rad/s ) 0.24 s = 4.124 rad/s