Question

A uniform solid sphere of radius r is placed at corner A of a table and is given a slight clockwise motion.Assuming that corner is sharp and motion.Assuming that corner is sharp and becomes slightly embedded in the sphere so that the coefficient of static friction at A is very large. qquad A| the corresponding velocity of the centre of the sphere.​

Answer 1

Answer:

Part i)

Angle made by the sphere from vertical is

$\beta = cos^{-1}(\frac{10}{17})$

Part ii)

velocity of the center of the sphere is

$v = \sqrt{\frac{10Rg}{17}}$

Explanation:

As we know that the sphere will rotate about the contact point on the table

So here we can say that the Normal force on the sphere will be ZERO when it is just about to drop

so we have

$mgcos\beta = \frac{mv^2}{R}$

$gcos\beta = \frac{v^2}{R}$

also by energy conservation

$mgR(1 - cos\beta) = \frac{1}{2}I\omega^2$

here we know

$I = \frac{7}{5}mR^2$

$\omega = \frac{v}{R}$

$mgR(1 - cos\beta) = \frac{1}{2}(\frac{7}{5}mR^2)(\frac{v^2}{R^2})$

$gR(1 - cos\beta) = \frac{7}{10}v^2$

now from above two equations we have

$g(1 - cos\beta) = \frac{7}{10}(gcos\beta)$

$1 - cos\beta = \frac{7}{10}cos\beta$

$1 = \frac{17}{10}cos\beta$

$cos\beta = \frac{10}{17}$

$\beta = cos^{-1}(\frac{10}{17})$

now from above equation the velocity of the COM of the sphere is given as

$v = \sqrt{Rg cos\beta}$

$v = \sqrt{Rg\times \frac{10}{17}}$

$v = \sqrt{\frac{10Rg}{17}}$

#Learn

Topic : Rotational Motion

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Answer 2

Answer:

the up answer is wrong if anyone could pls tell