(C) For the block to remain stationary on the rotating disc,the centripetal force must be provided by the static friction force.Let $m$ be the mass of

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(C) For the block to remain stationary on the rotating disc,the centripetal force must be provided by the static friction force.Let $m$ be the mass of the block,$\omega$ be the angular frequency,and $x$ be the distance from the axis.The condition for no sliding is: $f_s \leq \mu_s N$.Since $N = mg$ and $f_s = m\omega^2 x$,we have $m\omega^2 x \leq \mu_s mg$.Thus,$x \leq \frac{\mu_s g}{\omega^2}$.Given $\mu_s = 0.4$,$g = 10\, m/s^2$,and $\omega = 10\, rad/s$:$x = \frac{0.4 \times 10}{10^2} = \frac{4}{100}\, m$.Converting to centimeters: $x = 0.04\, m \times 100 = 4\, cm$.

Class 11 Physics Newton's Laws of Motion and Friction Circular motion with Friction

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$A$ disc of radius $5\, m$ is rotating with an angular frequency of $10\, rad/s$. $A$ block of mass $2\, kg$ is to be placed on the disc. If the coefficient of friction between the disc and the block is $\mu_s = 0.4$,find the maximum distance from the axis where the block can be placed without sliding (in $cm$).

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A
$2$
B
$3$
C
$4$
D
$6$

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Newton's Laws of Motion and Friction

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Circular motion with Friction

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