Explanation
Two vectors $$\overline{A}$$ and $$\overline{B}$$ lie in a plane. Another vector $$\overline{C}$$ lies outside this plane, then the resultant of the three vectors $$(\overline{A}+\overline{B}+\overline{C})$$ :
The observer will see only four spokes if the difference of angular
displacements of wheel 1 \& 2 is
$$n\dfrac { 2\pi }{ 4 } =n\dfrac { \pi }{ 2 } \\$$ Difference in angular displacement = (8rev/min - 6rev/min)\times t $$= 2\times 2\pi \times t=4\pi \times t (t in minutes)\\$$ The least time after which observer will see only four spokes is given by
$$4\pi \times t = (1)\frac { \pi }{ 2 } \\ $$ \therefore t= 1/8 min Therefore no of times observer will see only four spokes in a min= 8We can also think of this in following way: No of times observer will see only four spokes in a time period will be the value of n as it will tell us the number of $$(\dfrac { \pi }{ 2 } )$$ degrees rotations that the wheels have
taken relative to each other. And everytime they takes a $$\frac { \pi }{ 2 } $$ degree rotation relative to each other
observer will see only four spokes once.
In 1 min $$2\times 2\pi =4\pi =8\frac { \pi }{ 2 }$$ . Therefore 8 times a min.
Please disable the adBlock and continue. Thank you.