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Grade: 11
        
Can’t understand how to proceed with the above question 
10 months ago

Answers : (1)

Vikas TU
11680 Points
							
Dear student 
We know that the time period of an object,
T = 2π√(I/mgL)
where ,
I  = moment of inertia from the suspended point
L = distance of its centre from suspended point = r
we know that, the moment of inertia of disc about its centre = mr²/2
using parallel axis theoram the moment of inertia from a point in its periphery,
I = mr² + mr²/2 = 3mr²/2
putting the values in the above equation we get,
T = 2π√(3mr²/2mgr)
= 2π√(3r/2g)
Hence the time period of the disc is 2π√(3r/2g)
10 months ago
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