Question

Six identical square plates are welded to form a hollow cubical box of dimension 10cm.The total mass of the box is 1kg.The box is rotated about an axis passing through its geometrical center and perpendicular to two of the faces.Calculate the moment of inertia​

Answer 1

Answer:

  (2/9)*10^-2 kgm^2

Explanation: Dear student

four plates are equidistant to each other from geometrical center,and remaining two plates are parallel.

See the attachment for further process.

Answer 2

Given:

Total mass of box, $M=1 kg$

Dimension of cubical box, $a=10 cm=0.10 m$

Formula Used:

Moment of inertia, $I=mr^2$

where, m is the mass and r is the distance from the rotational axis.

Calculations:

Considering center of mass of each plate to lie at the center of each plate. Then, the moment of inertia of two plates through which the axis passes would be zero because the center of mass of the plate lies on the rotational axis making r = 0.

For rest of the four plates, the distance from the rotational axis is $\frac{a}{2}$.

Mass of each plate: $\frac{M}{6}$

So, moment of inertia would be:

$I=4\times \frac {M}{6}\frac {a^2}{2^2}$

Substitute the values and solve:

$I=\frac{1kg\times (0.10m)^2}{6}=0.00167 kg-m^2$

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