Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A boy is sitting on the horizontal platform of a joy wheel at a distance of $5\, m$ from the center. The wheel begins to rotate and when the angular speed exceeds $1\, rad / s$, the boy just slips. The coefficient of friction between the boy and the wheel is $\left(g=10\, m / s ^{2}\right)$

Laws of Motion

Solution:

$\frac{v^{2}}{r}=\omega^{2} r=\mu g$
$m=0.5$