(C) For a body rolling down an inclined plane without slipping,the condition for static friction is given by $\mu_s \geq \frac{\tan \theta}{1 + R^2/K^

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(C) For a body rolling down an inclined plane without slipping,the condition for static friction is given by $\mu_s \geq \frac{\tan \theta}{1 + R^2/K^2}$.For a solid cylinder,the radius of gyration $K$ is given by $K^2 = \frac{1}{2} R^2$,which implies $\frac{K^2}{R^2} = \frac{1}{2}$ or $\frac{R^2}{K^2} = 2$.Substituting this into the condition: $\mu_s \geq \frac{\tan \theta}{1 + 2}$.$\mu_s \geq \frac{\tan \theta}{3}$.Rearranging the terms,we get $\tan \theta \leq 3 \mu_s$.

Class 11 Physics System of Particles and Rotational Motion Rolling On Inclined Plane

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$A$ solid cylinder is rolling down on an inclined plane of angle $\theta$. The coefficient of static friction between the plane and the cylinder is $\mu_s$. The condition for the cylinder not to slip is

A
$\tan \theta \geq 3 \mu_s$
B
$\tan \theta > 3 \mu_s$
C
$\tan \theta \leq 3 \mu_s$
D
$\tan \theta < 3 \mu_s$

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System of Particles and Rotational Motion

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Rolling On Inclined Plane

$A$ circular disc reaches from top to bottom of an inclined plane of length $L$. When it slips down the plane,it takes time $t_{1}$. When it rolls down the plane,it takes time $t_{2}$. The value of $\frac{t_{2}}{t_{1}}$ is $\sqrt{\frac{3}{x}}$. The value of $x$ will be .... .

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The moment of inertia of a solid cylinder about its axis is $I$. It is allowed to roll down an incline without slipping. If its angular velocity at the bottom is $\omega$,then the total kinetic energy $(K.E.)$ of the cylinder will be:

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$A$ solid cylinder is rolling down on an inclined plane of angle $\theta$. The coefficient of static friction between the plane and the cylinder is $\mu_s$. The condition for the cylinder not to slip is

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$A$ circular disc reaches from top to bottom of an inclined plane of length $L$. When it slips down the plane,it takes time $t_{1}$. When it rolls down the plane,it takes time $t_{2}$. The value of $\frac{t_{2}}{t_{1}}$ is $\sqrt{\frac{3}{x}}$. The value of $x$ will be .... .

$A$ solid cylinder rolls down an inclined plane from a height $h$. At any moment,the ratio of rotational kinetic energy to the total kinetic energy is:

$A$ solid cylinder of mass $m$ and radius $R$ rolls down an inclined plane of height $30 \ m$ without slipping. The speed of its centre of mass when the cylinder reaches the bottom is $[$use $g=10 \ m \ s^{-2}]$ (in $m \ s^{-1}$)

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