Circumference 7209

The speed of the points lying on the circumference of the rotating disk is 6 m/s. The speed of the points, which lie 20 cm closer to the axis of rotation, is 4 m/s. Find the angular velocity of the wheel.

Correct answer:

o =  10 1/s

Step-by-step explanation:

$$\begin{gathered} v_1=6\text{m/s} \\ {v}_2=4\text{m/s} \\ d=20\text{cm}\to \text{m}=20:100\text{m}=0.2\text{m}; \\ v=\omega \cdot r \\ {v}_1=o\cdot r \\ {v}_2=o\cdot (r-d) \\ {v}_2=v_1/r\cdot (r-d) \\ {v}_2\cdot r=v_1\cdot (r-d) \\ {v}_2\cdot r=v_1\cdot r-v_1\cdot d \\ r=\frac{v_1\cdot d}{v_1-v_2}=\frac{6\cdot \frac{1}{5}}{6-4}=\frac{3}{5}=0.6\text{m} \\ o=v_1/r=6/\frac{3}{5}=6:\frac{3}{5}=6\cdot \frac{5}{3}=\frac{6\cdot 5}{3}=\frac{30}{3}=101\text{/s} \end{gathered}$$

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