Question

In Fig.$10-31$ , wheel A of radius ${\mathit{r}}_{\mathbf{A}}\mathbf{=}\mathbf{}\mathbf{10}\mathbf{ }\mathbf{\text{cm}}$is coupled by belt B to wheel C of radius $r_C=25 \text{cm}$ .The angular speed of wheel A is increased from rest at a constant rate of$\mathbf{1}\mathbf{.}\mathbf{6}\raisebox{1ex}{${\displaystyle \mathbf{\text{rad}}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\mathbf{\text{s}}}^{\mathbf{\text{2}}}}$}\right.$ . Find the time needed for wheel C to reach an angular speed of $100\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.$ , assuming the belt does not slip. (Hint: If the belt does not slip, the linear speeds at the two rims must be equal.)

Short answer

The time needed for wheel C to reach an angular speed of $100\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.$,$t\text{is}16 \text{s}$ .

Step-by-step solution

Understanding the given information

1. The radius of wheel A, $r_A$is $0.10 \text{m}$.
2. The radius of wheel C,$r_C$is$0.25 \text{m}$.
3. The angular acceleration of wheel A,${\alpha }_A$is$1.6 \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$.
4. The angular speed $\omega$ is $100 \raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.$.

Concept and Formula used for the given question

By using the formulas for tangential acceleration and angular speed, we can find thetime needed for wheel C to reach an angular speed of

1. The tangential acceleration $a_t=\alpha r$
2. The angular speed $\omega$is$\omega =\alpha t$
3. Hint: If the belt does not slip, the linear speed at the two rims must be equal.

Calculation for the time needed for wheel C to reach an angular speed of  100revmin

Since the belt does not slip, a point on the rim of wheel C has the same tangential acceleration as a point on rim of wheel A. This means that

${\alpha }_A{r}_A={\alpha }_C{r}_C$

Where, ${\alpha }_A$ is the angular acceleration of wheel A, and ${\alpha }_C$is the angular acceleration of wheel C. Thus,

${\alpha }_C=\frac{r_A}{r_C}{\alpha }_A$

Substitute all the value in the above equation.

$$\begin{gathered} {\alpha }_C=\left(\frac{0.10 \text{m}}{0.25 \text{m}}\right)\times 1.6\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right. \\ =0.64\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right. \end{gathered}$$

Angular speed of wheel C given by

${\omega }_C={\alpha }_{C}t$

The time t for it to reach an angular speed$\omega =100 \raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{min}}$}\right.=10.5\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$starting from rest is given by

$t=\frac{{\omega }_C}{{\alpha }_C}$

Substitute all the value in the above equation.

$$\begin{gathered} t=\frac{{\displaystyle 10.5\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{{\displaystyle 0.64 \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}} \\ =16 \text{s} \end{gathered}$$

Hence the time is,$16 \text{s}$ .