In figure, wheel A of radius ${\mathrm{r}}_{\mathrm{A}}$ = 10 cm is coupled by belt B to wheel C of radius ${\mathrm{r}}_{\mathrm{C}} =$25 cm. The angular speed of wheel A is increased from rest at a constant rate of 1.6 $\mathrm{rad}/{\mathrm{s}}^2$. Find the time needed for wheel C to reach an angular speed of 12.8 rad/s, assuming the belt does not slip.

If the belt does not slip, the linear speeds at the two rims must be equal. Since the belt does not slip, a point on the rim of wheel C has the same tangential acceleration as a point on the rim of wheel A. This means that ${\text{α}}_{\mathrm{A}}{\mathrm{r}}_{\mathrm{A}} = {\mathrm{\alpha }}_{\mathrm{C}}{\mathrm{r}}_{\mathrm{C}}$where ${\mathrm{\alpha }}_{\mathrm{A}}$ is the angular acceleration of wheel A and ${\mathrm{\alpha }}_{\mathrm{C}}$ is the angular acceleration of wheel C. Thus,

${\mathrm{\alpha }}_{\mathrm{C}} = \left(\frac{{\mathrm{r}}_{\mathrm{A}}}{{\mathrm{r}}_{\mathrm{C}}}\right){\mathrm{\alpha }}_{\mathrm{C}} = \left(\frac{10 \mathrm{cm}}{25 \mathrm{cm}}\right)(1.6 \mathrm{rad}/{\mathrm{s}}^2) = 0.64 \mathrm{rad}/{\mathrm{s}}^2.$

With the angular speed of wheel C given by ${\mathrm{\omega }}_{\mathrm{C}} = {\mathrm{\alpha }}_{\mathrm{C}}\mathrm{t}$, the time for it to reach an angular speed of $\mathrm{\omega }$ = 100 rev/min = 10.5 rad/s starting from rest is

$\mathrm{t} = \frac{{\mathrm{\omega }}_{\mathrm{C}}}{{\mathrm{\alpha }}_{\mathrm{C}}} = \frac{12.8 \mathrm{rad}/\mathrm{s}}{0.64 \mathrm{rad}/{\mathrm{s}}^2} = 20 \mathrm{s}$