Explanation:
We can solve this problem by applying the work-energy theorem, which states that the work done by all the forces acting on an object W results in the change in its kinetic energy. This can be stated mathematically as follows:
[tex]W = \Delta{K} = K- K_0[/tex]
where [tex]K_0[/tex] is the initial kinetic energy and [tex]K[/tex] is the final kinetic energy. Since the merry-go-round started from rest, we know that its initial kinetic energy is zero, thus our work-energy theorem reduces to
[tex]W = K[/tex]
Moreover, since we are dealing with a rotating object (i.e., the merry-go-round), then we need to use an expression for the kinetic energy that is more suited to rotational motion. Hence, our work-energy theorem becomes
[tex]W = \frac{1}{2}I\omega^2[/tex]
where I is the moment of inertia, which depends on the mass and shape of the object and is measured in [tex]\text{kg-m}^2[/tex] and [tex]\omega[/tex] is the angular velocity measured in rad/s.
We are given the following quantities:
[tex]m = 1640\:\text{kg}[/tex]
[tex]r = 7.5\:\text{m}[/tex]
[tex]\theta = 3\:\text{rev}[/tex]
[tex]t = 24\:\text{s}[/tex]
Let's solve for the angular velocity first. Since the merry-go-round takes 24 seconds to complete 3 revolutions, its angular velocity is
[tex]\omega = \dfrac{3\:\text{rev}}{24\:\text{s}} = 0.125\:\text{rev/s}[/tex]
Since angular velocity is measured in rad/s, we need to convert rev/s to rad/s:
[tex]0.125\:\dfrac{\text{rev}}{\text{s}}×\dfrac{2\pi\:\text{rad}}{1\:\text{rev}} = 0.785\:\text{rad/s}[/tex]
Therefore, our angular velocity [tex]\omega[/tex] is
[tex]\omega = 0.785\:\text{rad/s}[/tex]
The next thing we need to do is to find the moment of inertia of the merry-go-round. Since we can treat it like a solid cylinder, the moment of inertia for such a shape is given by
[tex]I = \frac{1}{2}mr^2[/tex]
where m and r are the mass and the radius of the merry-go-round, respectively. Plugging in the given values, we find that the moment of inertia is
[tex]I = \frac{1}{2}(1640\:\text{kg})(7.5\:\text{m})^2 = 46,125\:\text{kg-m}^2[/tex]
Using all the values that we have calculated so far, the work done in accelerating the merry-go-round from rest to 0.785 rad/s is
[tex]W = \frac{1}{2}(46125\:\text{kg-m}^2)(0.785\:\text{rad/s})^2[/tex]
[tex]\:\:\:\:\:\:= 14,212\:\text{Joules}[/tex]
