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Question
A big dumb-bell is prepared by using a uniform rod of mass 60 g and length 20 cm. Two identical solid spheres of mass 25 g and radius 10 cm each are at the two ends of the rod. Calculate the moment of inertia of the dumb-bell when rotated about an axis passing through its centre and perpendicular to the length.
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Solution
Given:
Msph = 25 g,
Rsph = 10 cm,
Mrod = 60 g,
Lrod = 20 cm
To find:
moment of inertia of the dumb-bell
Solution:
The MI of a solid sphere about its diameter is
`I_("sph","CM") = 2/5 M_"sph"R_"sph"^2`
The distance of the rotation axis (transverse symmetry axis of the dumbbell) from the centre of the sphere is h = 20cm.
The MI of a solid sphere about the rotation axis,
`I_"sph" = I_("sph","CM") + M_"sph"h^2`
For the rod, the rotation axis is its transverse symmetry axis through CM.
The MI of a rod about this axis,
Irod = `1/12 M_"rod" L_"rod"^2`
Since there are two solid spheres, the MI of the dumbbell about the rotation axis is
i = 2Isph + Irod
`= 2 M_"sph"(2/5 R_"sph"^2 + h^2) + 1/12 M_"rod" L_"rod"^2`
= `2(25)[2/5(10)^2 + (20)^2] + 1/12(60)(20)^2`
`= 50 [2/5 (100) + 400] + 1/12(60)(400)`
`= 50 [2/cancel(5) xx cancel(100)^20 + 400] + 1/cancel(12) xx cancel(60)^5 xx (400)`
= 50 [2 × 20 + 400] + 5 × 400
= 50 × 440 + 2000
= 22000 + 2000
= 24000 g cm2
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