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Standard XII

Physics

Spring Potential Energy

From a semici...

Question

From a semicircular disc of mass m and radius R a circular portion of radius $\frac{R}{2}$ is removed as shown in figure then moment of inertia of remaining portion about diameter of semicircle as shown in figure is $\frac{m R^{2}}{n}$ . Here n is (Write upto two digits after the decimal point.)

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Solution

$I = I_{1} + I_{2} \Rightarrow I_{1} = I - I_{2}$

$I = \frac{m R^{2}}{4}$ and $I_{2} = \frac{m_{2} {(\frac{R}{2})}^{2}}{4} + m_{2} {(\frac{R}{2})}^{2}$

where, $m_{2} = \frac{M}{2}, I_{2} = \frac{5 m R^{2}}{32}$

$I_{1} = \frac{m R^{2}}{4} - \frac{5 m R^{2}}{32} = \frac{3}{32} m R^{2}$

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From a semicircular disc of mass m and radius R a circular portion of radius R2 is removed as shown in figure then moment of inertia of remaining portion about diameter of semicircle as shown in figure is mR2n. Here n is (Write upto two digits after the decimal point.)

I=I1+I2⇒I1=I−I2 I=mR24 and I2=m2(R2)24+m2(R2)2 where, m2=M2, I2=5mR232 I1=mR24−5mR232=332mR2

From a semicircular disc of mass m and radius R a circular portion of radius $\frac{R}{2}$ is removed as shown in figure then moment of inertia of remaining portion about diameter of semicircle as shown in figure is $\frac{m R^{2}}{n}$ . Here n is (Write upto two digits after the decimal point.)

$I = I_{1} + I_{2} \Rightarrow I_{1} = I - I_{2}$

$I = \frac{m R^{2}}{4}$ and $I_{2} = \frac{m_{2} {(\frac{R}{2})}^{2}}{4} + m_{2} {(\frac{R}{2})}^{2}$

where, $m_{2} = \frac{M}{2}, I_{2} = \frac{5 m R^{2}}{32}$

$I_{1} = \frac{m R^{2}}{4} - \frac{5 m R^{2}}{32} = \frac{3}{32} m R^{2}$