For a short period of time, the motor turns gear A with a constant angular acceleration of...

Initially gear is at rest.

For gears \(A\) and \(B\), we have \(r_{A} \alpha_{A}=r_{B} \alpha_{B}\)

Accordingly,

$$ \begin{aligned} &0.075 \times 4.5=0.225 \times \alpha_{B} \\ &0.225 \alpha_{B}=0.3375 \\ &\alpha_{B}=1.5 \mathrm{rad} / \mathrm{s}^{2} \end{aligned} $$

Angular velocity of gear \(B\) is expressed as

$$ \omega_{B}=\left(\omega_{B}\right)_{0}+\alpha_{B} t $$

Accordingly, we have

$$ \begin{aligned} \omega_{B} &=0+1.5 \times 3 \\ \omega_{B} &=4.5 \mathrm{rad} / \mathrm{s} \end{aligned} $$

We also have angular displacement expressed as

$$ \theta_{B}=\left(\theta_{B}\right)_{0}+\left(\omega_{B}\right)_{0} t+\frac{1}{2} \alpha_{B} t^{2} $$

By substituting the values of the parameters into the above relation, we have

$$ \begin{aligned} &\theta_{B}=0+(0 \times 3)+\left(\frac{1}{2} \times 1.5 \times 3^{2}\right) \\ &\theta_{B}=6.75 \mathrm{rad} \end{aligned} $$

Velocity of the cylinder is expressed as

$$ v_{C}=r_{D} \omega_{B} $$

Accordingly, we have

$$ \begin{aligned} &v_{C}=0.125 \times 4.5 \\ &v_{C}=0.5625 \mathrm{~m} / \mathrm{s} \end{aligned} $$

Distance travelled by the cylinder is expressed as

$$ s_{C}=r_{D} \theta_{B} $$

Accordingly, we have

$$ \begin{aligned} &s_{C}=0.125 \times 6.75 \\ &s_{C}=0.844 \mathrm{~m} \end{aligned} $$

Distance travelled by the cylinder is \(s_{C}=0.844 \mathrm{~m}\)