Question

The angular velocity of a 755-g wheel 15.0 cm in diameter is given by the equation ω(t) = (2.00 rad/s2)t + (1.00 rad/s4)t 3. (a) Through how many radians does the wheel turn during the first 2.00 s of its motion? (b) What is the angular acceleration (in rad/s2) of the wheel at the end of the first 2.00 s of its motion?

Answer 1

Answer:

a

The number of radians turned by the wheel in 2s is   $\theta= 8\ radians$

b

The angular acceleration is  $\alpha =14 rad/s^2$

Explanation:

        The angular velocity  is given as

                 $w(t) = (2.00 \ rda/s^2)t + (1.00 rad /s^4)t^3$

Now generally the integral of angular velocity gives angular displacement

           So integrating the equation of angular velocity through the limit 0 to 2 will gives us the angular displacement for 2 sec

    This is mathematically evaluated as

            $\theta(t ) = \int\limits^2_0 {2t + t^3} \, dt$

                  $= [\frac{2t^2}{2} + \frac{t^4}{4}] \left\{ 2} \atop {0}} \right.$

                  $= [\frac{2(2^2)}{2} + \frac{2^4}{4}] - 0$

                  $= 4 +4$

                 $\theta= 8\ radians$

Now generally the derivative  of angular velocity gives angular acceleration

      So the value of the derivative of angular velocity equation at t= 2 gives us the angular acceleration

    This is mathematically evaluated as          

           $\frac{dw}{dt} = \alpha (t) = 2 + 3t^2$

so at t=2

            $\alpha (2) = 2 +3(2)^2$

                   $\alpha =14 rad/s^2$