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  2. Solve! A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R . Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m , if the string does not slip

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Solve! A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R . Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m , if the string does not slip

A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R . Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m , if the string does not slip on the pulley, is

Option 1)

\frac{3}{2}g

 Option 2)

g
Option 3)

\frac{2}{3}g

 Option 4)

\frac{g}{3}
Answers (1)
762 Views
A Aadil Khan

As we learnt in

Rolling of a body on an inclined plane -

a= \frac{g\sin \Theta }{1+\frac{K^{2}}{R^{2}}}

f= \frac{mg\sin \Theta }{1+\frac{R^{2}}{K^{2}}}

- wherein

K=Radius of gyration

\Theta = Angle of inclination

 

 

 

a=\frac{Mg}{m+\frac{I}{R^{2}}}=\frac{Mg}{m+\frac{1}{2}\frac{mR^{2}}{R^{2}}}

a=\frac{2Mg}{3m}

a=\frac{2g}{3}


Option 1)

\frac{3}{2}g

This is an incorrect option.

Option 2)

g

This is an incorrect option.

Option 3)

\frac{2}{3}g

This is the correct option.

Option 4)

\frac{g}{3}

This is an incorrect option.

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