Figure$\mathbf{\text{10-21}}$shows plots of angular position$\mathbf{\text{θ}}$versus time t for three cases in which a disk is rotated like a merry-go-round. In each case, the rotation direction changes at a certain angular position${\mathbf{\theta }}_{\mathrm{change}}$. (a) For each case, determine whether${\mathbf{\theta }}_{\mathrm{change}}$is clockwise or counterclockwise from$\mathbf{\theta }\mathbf{=}\mathbf{}\mathbf{0}$, or whether it is at$\mathbf{\theta }\mathbf{=}\mathbf{}\mathbf{0}$. For each case, determine (b) whether v is zero before, after, or at$\mathbf{t}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}$and (c) whether a is positive, negative or zero.

Figure 10-21 with 3 plots of $\mathrm{\theta }$ versust.

The rotation of the radius vector is called angular displacement. It is measured in radians. The rate of change of angular displacement with respect to time is called as angular velocity. It is measured in radians per second. The rate of change of angular velocity with respect to time is called angular acceleration. The rate of change of displacement of radius vector with respect to time is linear velocity.

From the graph and the concept of slope and vertex of curves, mark the situation for any graph.

Formulae are as follows:

$\mathrm{\omega }=\frac{\mathrm{d\theta }}{\mathrm{dt}}$

$\mathrm{v}=\mathrm{r\omega }$

${\mathrm{a}}_{\mathrm{t}\text{angential}}=\mathrm{r}\mathrm{\alpha }$

${\mathrm{a}}_{\text{radial}}={\mathrm{\omega }}^2\mathrm{r}$

Here,ω is the angular velocity, θis angular displacement, ris the radius, v is linear velocity, and α is angular acceleration.

As it is known, a change in the direction happens when the graph changes from negative to positive or vice versa. And the point of changing the direction on the curve is called a vertex.

If the point is above the x-axis, then${\mathrm{\theta }}_{\mathrm{change}}\text{is}$ counterclockwise. And if it is below the x-axis, it is clockwise. So,

For Curve 1,${\mathrm{\theta }}_{\mathrm{change}}$ occurs at counterclockwise from$\mathrm{\theta }=0$

For Curve 2,${\mathrm{\theta }}_{\mathrm{change}}$occurs at counterclockwise from$\mathrm{\theta }=0$

For Curve3, ${\mathrm{\theta }}_{\mathrm{change}}$occurs at$\mathrm{\theta }=0$

Angular velocity$\mathrm{\omega }$ is the slope of the $\mathrm{\theta }$vs time graph. So when the slope is zero, angular velocity is equal to zero. By observing the graph, determine the points at which the slope is zero.

For Curve 1, it is before

For Curve 2, it is at $\mathrm{t}=0$

And for Curve 3, it is after

In order to find out if alpha is positive, look at the second derivative of a $\mathrm{\theta }$vs $\mathrm{t}$ graph. Since Curve 1 and Curve 3 are concave up, angular acceleration is positive for them. On the other hand, the Curve 2 is concave down, so angular acceleration is negative for it.

So,

For Curve 1, angular acceleration is positive.

For Curve 2, angular acceleration is negative.

For Curve 3, angular acceleration is positive.

Therefore, Using the given graph and the vertex point, and the slope of the curve, all the desired values can be found.