A disc rotates about its axis of symmetry in | vedclass

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Question

$\begin{array}{l}
{m{Using,}}\,\mu {m{mg = }}\frac{{m{v^2}}}{r} = mr{\omega ^2}\
\omega  = 2\pi n = 2\pi  	imes 3.5 = 7\pi rad/\sec

Vedclass · Accepted Answer

$0.6$

Vedclass · Answer

$\begin{array}{l}
{m{Using,}}\,\mu {m{mg = }}\frac{{m{v^2}}}{r} = mr{\omega ^2}\
\omega  = 2\pi n = 2\pi  	imes 3.5 = 7\pi rad/\sec \
Radius,\,r = 1.25\,cm = 1.25 	imes {10^{ - 2}}m\
Coefficient\,of\,friction,\,\mu  = ?\
\mu mg = \frac{{m{{\left( {r\omega } ight)}^2}}}{r}\left( {v = r\omega } ight)
\end{array}$

$\begin{array}{l}
 \Rightarrow \mu  = \frac{{r{\omega ^2}}}{g} = \frac{{1.25 	imes {{10}^{ - 2}} 	imes {{\left( {7 	imes \frac{{22}}{7}} ight)}^2}}}{{10}}\
\,\,\,\,\,\,\,\, = \frac{{1.25 	imes {{10}^{ - 2}} 	imes {{22}^2}}}{{10}} = 0.6
\end{array}$

A disc rotates about its axis of symmetry in a hoizontal plane at a steady rate of $3.5$ revolutions per second. A coin placed at a distance of $1.25\,cm$ from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is $(g\, = 10\,m/s^2)$

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