Question #51386

The moment of inertia of a annular disc of mass M,the inner diameter R1 and outer diameter is R2 about perpendicular to the axis is
1

Expert's answer

2015-03-17T13:18:03-0400

Answer on Question #51386-Physics-Mechanics-Kinematics-Dynamics

The moment of inertia of a annular disc of mass MM , the inner diameter R1R_{1} and outer diameter is R2R_{2} about a perpendicular axis through its center is

Solution


An annular disc is just a circular disc from which a smaller, concentric disc has been removed, leaving a hole in its place as shown in the figure.

Let OO be the center of annular disc. Then, clearly, face area of the disc is π(R22R12)\pi (R_2^2 - R_1^2) and, therefore,

mass per unit area of the disk =Mπ(R22R12)= \frac{M}{\pi\left(R_2^2 - R_1^2\right)}

Now, consider a coaxial ring of radius xx and width dxdx shown in the figure. We clearly have,

face area of the ring =2πxdx= 2\pi xdx and, therefore, it mass =Mπ(R22R12)2πxdx=M2x(R22R12)dx.= \frac{M}{\pi(R_2^2 - R_1^2)} 2\pi xdx = \frac{M2x}{(R_2^2 - R_1^2)} dx.

Hence, the moment of inertia of this ring about an axis through OO and perpendicular to its plane is


M2x(R22R12)dxx2=M2x3(R22R12)dx.\frac {M 2 x}{\left(R _ {2} ^ {2} - R _ {1} ^ {2}\right)} d x \cdot x ^ {2} = \frac {M 2 x ^ {3}}{\left(R _ {2} ^ {2} - R _ {1} ^ {2}\right)} d x.


The moment of inertia of the whole annular disk about an axis through O and perpendicular to its plane is thus easily obtained by integrating the above expression for the moment of inertia of this ring, between the limits x=R1x = R_{1} and x=R2x = R_{2} , so that, we have


I=R1R2M2x3(R22R12)dx=2M(R22R12)R1R2x3dx=2M(R22R12)[x44]R1R2=2M(R22R12)[(R24R14)4]=M(R22+R12)2.\begin{array}{l} I = \int_ {R _ {1}} ^ {R _ {2}} \frac {M 2 x ^ {3}}{\left(R _ {2} ^ {2} - R _ {1} ^ {2}\right)} d x = \frac {2 M}{\left(R _ {2} ^ {2} - R _ {1} ^ {2}\right)} \int_ {R _ {1}} ^ {R _ {2}} x ^ {3} d x = \frac {2 M}{\left(R _ {2} ^ {2} - R _ {1} ^ {2}\right)} \left[ \frac {x ^ {4}}{4} \right] _ {R _ {1}} ^ {R _ {2}} = \frac {2 M}{\left(R _ {2} ^ {2} - R _ {1} ^ {2}\right)} \left[ \frac {\left(R _ {2} ^ {4} - R _ {1} ^ {4}\right)}{4} \right] \\ = \frac {M (R _ {2} ^ {2} + R _ {1} ^ {2})}{2}. \\ \end{array}


Answer: M(R22+R12)2\frac{M(R_2^2 + R_1^2)}{2} .

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