HSC Science (General) 12th Board ExamMaharashtra State Board
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State an Expression for the Moment of Intertia of a Solid Uniform Disc, Rotating About an Axis Passing Through Its Centre, Perpendicular to Its Plane. - HSC Science (General) 12th Board Exam - Physics

ConceptTheorems of Perpendicular and Parallel Axes

Question

State an expression for the moment of intertia of a solid uniform disc, rotating about an axis passing through its centre, perpendicular to its plane. Hence derive an expression for the moment of inertia and radius of gyration:

i. about a tangent in the plane of the disc, and

ii. about a tangent perpendicular to the plane of the disc.

Solution

The M. I of a thin uniform disc about an axis passing through its centre and perpendicular to its plane is given by,

IC = (1/2)MR2

i. According to theorem of parallel axis,

IT = Id + Mh2 = Id + MR2 [∵ h = R]

But Id = MR2/4

therefore I_T=(MR^2)/4+MR^2

therefore I_T=5/4 MR^2

Now , radius of gyration is given by,

K=sqrt(I/M)

therefore K=sqrt((5MR^2)/(4M))

thereforeK=sqrt5/2R

ii. Applying theorem of parallel axis,

IT = IO + Mh2 = IO + MR2 [∵ h = R]

But IO = MR2/2

I_T=(MR^2)/2+MR^2=3/2MR^2

Now, radius of gyration is given by,

K=sqrt(I/M)

thereforeK=sqrt((3MR^2)/(2M))

thereforeK=sqrt(3/2)R

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APPEARS IN

2014-2015 (October) (with solutions)
Question 1.1 | 7.00 marks
Solution State an Expression for the Moment of Intertia of a Solid Uniform Disc, Rotating About an Axis Passing Through Its Centre, Perpendicular to Its Plane. Concept: Theorems of Perpendicular and Parallel Axes.
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