91Ó°ÊÓ

The bicycle wheel is rotating at $50\text{rpm}$, and when the cyclist begins to pedal harder it gives the wheel a constant angular acceleration of$0.50\text{rad}/{\text{s}}^2$.

The angular velocity of the wheel after $10\text{s}$is,

${\mathrm{Ó\neg }}_{10\text{s}}={\mathrm{Ó\neg }}_i+\mathrm{Î\pm }t$, ${\mathrm{Ó\neg }}_{10\text{s}}$is the angular velocity of the wheel after $10\text{s}$, ${\mathrm{Ó\neg }}_i$is the initial angular velocity of the wheel, $\mathrm{Î\pm }$is the angular acceleration of the wheel, $t$is the period of time.

Substitute$50\text{rpm}$for${\mathrm{Ó\neg }}_i,0.50\text{rad}/{\text{s}}^2$for$\mathrm{Î\pm }$and$10\text{s}$for$t$in the above equation to find ${\mathrm{Ó\neg }}_{10\text{s}}$.

${\mathrm{Ó\neg }}_{10\text{s}}=\left(50\text{rpm}\right)+\left(0.50\text{rad}/{\text{s}}^2×\frac{1\text{rev}}{2\mathrm{Ï€}\text{rad}}×\frac{60\text{s}}{1\text{min}}\right)(10\text{s})$

$\begin{array}{l}=\left(50\text{rpm}\right)+47.74\text{rpm}\\ =97.74\text{rpm}\end{array}$

Thus, the angular velocity of the wheel after$10\text{s}$ is$97.74\text{rpm}$.

Given info: The bicycle wheel is rotating at $50\text{rpm}$, and when the cyclist begins to pedal harder it gives the wheel a constant angular acceleration of $0.50\text{rad}/{\text{s}}^2$.

The revolution made by the wheel in $10\text{s}$is,$\mathrm{θ}=2\mathrm{π}N…….\left(\text{l}\right)$

Here, $\mathrm{θ}$is the angular displacement in $10\text{s}$, and $N$is the number of revolution made by the wheel in $10\text{s}$.

The angular displacement made by the wheel in $10\text{s}$is, $\mathrm{θ}={\mathrm{Ó\neg }}_{i}t+\frac{1}{2}\mathrm{Î\pm }{t}^2$

Here, ${\mathrm{Ó\neg }}_i$is the initial angular velocity of the wheel, $\mathrm{Î\pm }$is the angular acceleration of the wheel, and $t$is the period of time.

$\mathrm{θ}=\left(50\text{rpm}×\frac{1\text{min}}{60\text{s}}\right){(10\text{s})+\frac{1}{2}\left(0.50\text{rad}/{\text{s}}^2\right)10\text{s})}^2$

$\begin{array}{l}=52.33\text{rad}+25\text{rad}\\ =77.33\text{rad}\end{array}$

Thus, the angular displacement of the wheel in $10\text{s}$is $77.33\text{rad}$.

Substitute $77.33\mathrm{rad}$for $\mathrm{θ}$in the equation (I) to find $N$.

$77.33\text{rad}=2\mathrm{Ï€}N$$N=\frac{77.33\text{rad}}{2\mathrm{Ï€}}=12.30\text{rev}$

Therefore, the number of revolution made by the wheel in $10\text{s}$is $12.30\text{rev}$.