As per the question,
Acceleration in the rim of the wheel is "(a)=9.8 m\/sec^2"
Radius of the wheel (R)="1.0 \\times 10^2 m"
We know that "a=R\\omega^2"
"\\omega=" angular velocity of the wheel
We know that angular velocity "(\\omega)=\\dfrac{2\\pi}{T}"
where T is the time period.
So,
"9.8 m\/sec^2=1.0\\times \\dfrac{2\\pi}{T}"
"\\Rightarrow T=\\dfrac{2\\pi}{9.8}=0.64 sec"
or 0.6 sec
Comments
The answer is wrong as the author forgot to perform the square root. the correct answer will be 20 s.
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