Moment of Inertia (M.I.) of four bodies having same mass 'M' and radius '2R' are as follows:
I1 = M.I. of solid sphere about its diameter
I2 = M.I. of solid cylinder about its axis
I3 = M.I. of solid circular disc about its diameter
I4 = M.I. of thin circular ring about its diameter
If 2(I2 + I3) + I4 = x . I1, then the value of x will be __________.
Answer (integer)
5
Solution
<p>$$2\left( {{1 \over 2} + {1 \over 4}} \right) \times M{(2R)^2} + {1 \over 2}M{(2R)^2} = x{2 \over 5}M{(2R)^2}$$</p>
<p>$\Rightarrow 1 + {1 \over 2} + {1 \over 2} = x \times {2 \over 5}$</p>
<p>$\Rightarrow x = 5$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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