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Question

Physics Question on Moment Of Inertia

The moment of inertia of a solid flywheel about its axis is 0.1kgm20.1 \,kg\, m^2. A tangential force of 2 kg wt. is applied round the circumference of the flywheel with the help of a string and mass arrangement as shown in the figure. If the radius of the wheel is 0.1 m, find the acceleration of the mass

A

163.3rads2163.3\, rad\, s^{-2}

B

16.3rads216.3\, rad\, s^{-2}

C

81.66rads281.66\, rad\, s^{-2}

D

8.16rads28.16\, rad\, s^{-2}

Answer

16.3rads216.3\, rad\, s^{-2}

Explanation

Solution

Suppose a be the linear acceleration of the mass and T the tension in the string. Hence, MgT=Ma...(i)Mg - T = Ma \,...(i) Let ?? be the angular acceleration of the flywheel. The couple applied to the flywheel is Iα=TRI\alpha=TR or T=IαR...(ii)T=\frac{I\alpha}{R}\,...\left(ii\right) Now we know that a=R?...(iii)a = R? \,...\left(iii\right) Putting (ii)\left(ii\right) and (iii)\left(iii\right) in e (i),\left(i\right), we get Mg=IαR=MRαMg=\frac{I\alpha}{R}=MR\alpha α=MgRI+MR2...(iv)\therefore \alpha=\frac{MgR}{I+MR^{2}}\,...\left(iv\right) =2×9.8×0.10.1+2×(0.1)2=16.33rads2=\frac{2\times9.8\times0.1}{0.1+2\times\left(0.1\right)^{2}}=16.33\,rad\,s^{-2}