Figure$\mathbf{\text{10-20}}$is a graph of the angular velocity versus time for a disk rotating like a merry-go-round. For a point on the disk rim, rank the instants a, b, c, and d according to the magnitude of the (a) tangential and (b) radial acceleration, greatest first.

Graph of angular velocity vs time for a disk rotating like a merry-go-round is given.

From the slope of the graph rank the points according to the tangential acceleration. Then using the relation between radial acceleration and angular velocity rank the points according to the radial acceleration.

Formulae are as follows:

${\mathrm{a}}_{\text{tangential}}=\mathrm{r}.\mathrm{\alpha }$

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

Whereω is the angular velocity, αis angular acceleration and ris the radius.

The formula for tangential acceleration is given as,

${\mathrm{a}}_{\text{tangential}}=\mathrm{r}.\mathrm{\alpha }$

From the graph, we can say that$\mathrm{\alpha }$ is the slope of the$\mathrm{\omega }$vs time graph. The radius of the disk is the same for all cases, so the value of tangential acceleration depends on angular acceleration. The slope is high at point c, so angular acceleration is maximum. After that point a has a high slope and points b and d have zero slope.

So the ranking is as,

$\text{c}>\text{a}>\text{b}=\text{d}$

The formula for radial acceleration is,

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

The radius of the disk is the same, so the radial acceleration depends on the value of $\mathrm{\omega }$ which can be easily obtained from the graph. So ranking is,

$\text{b}>\text{a}=\text{c}>\text{d}$

Therefore, Observing the graph of angular velocity vs time find the radial and tangential acceleration of an object.