Givens.
• Radius = 0.26 m.
,• Frequency = 51 rev/min.
The period is the inverse of the frequency
[tex]\begin{gathered} T=\frac{1}{f}=\frac{1}{51\cdot\frac{\text{rev}}{\min }} \\ T=0.02\cdot\frac{\min }{\text{rev}} \end{gathered}[/tex]Then, we transform it into seconds.
[tex]\begin{gathered} T=0.02\cdot\frac{\min}{\text{rev}}\cdot\frac{60\sec }{1\min } \\ T=1.2\cdot\frac{\sec }{rev} \end{gathered}[/tex]The period in seconds is 1.2.
Then, the tangential speed formula is
[tex]v=\omega\cdot r[/tex]Where omega represents the angular speed, which in this case can be found using the frequency.
[tex]\begin{gathered} v=2\pi f\cdot r \\ v=2\pi\cdot51\cdot\frac{\text{rev}}{\min}\cdot\frac{1\min}{60\sec}\cdot0.26m \\ v=1.39\cdot\frac{m}{s} \end{gathered}[/tex]Therefore, the tangential speed is 1.39 meters per second.
