Question

A tyre is prepared by using a
uniform rod of mass 60 g and length 20
cm. Two identical solid spheres of mass
50 g and radius 10 cm each are at the
two ends of the rod. Calculate moment
of inertia of the tyre when rotated
about an axis passing through its centre
and perpendicular to the length?​

Answer 1

Answer:

2.4

Explanation:

60×20=1200

50×10=500

1200÷500

=2.4

Answer 2

Answer:A big tyre is prepared by using a uniform rod of mass 60 g and length 20cm. Two identical solid spheres of mass 50 g and radius 10 cm each are at the two ends of the rod.

To find:

Moment Of Inertia when rotated along an axis passing through the centre of system and perpendicular to its length.

Calculation:

Moment of inertia is actually the rotational analogue of mass and hence can be calculated by algebraic addition of individual moment of inertia of the different structures.

Total Moment Of Inertia

= MI(rod) +2 \times  MI(sphere)

=  \dfrac{(m1) {l}^{2} }{12}  +  \bigg \{2 \times  \bigg( \dfrac{2}{5} m {r}^{2}  + m {d}^{2}  \bigg) \bigg \}

=  \dfrac{(0.06) {(0.2)}^{2} }{12}  +  \bigg \{2 \times  \bigg( \dfrac{2}{5}  \times 0.05 \times  {(0.1)}^{2}  + 0.05 \times  {(0.2)}^{2}  \bigg) \bigg \}

= 0.0002 + 2 \times (0.0022m)

= 0.0002 + 0.0044

=0.0046 \: kg  \: {m}^{2}  

So , final answer is :

Moment Of Inertia is 0.0046 kg m².

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So , final answer is :

Moment Of Inertia is 0.0046 kg m².