Physics, asked by yuvagokul28, 1 month ago

find the moment of inertia with a disc from axis perpendicular to the center using integration method. ​

Answers

Answered by IamIronMan0
62

Answer:

Let mass density of disk be p ( rho ) , mass m and radius R .

So

M =  \rho.\pi {R}^{2}

Now mass of small ring of mass dm radius r and dr bridth is

dM =  \rho (2\pi \: rdr)

So moment of inertia

dI= (dM) {r}^{2}  =( \rho. 2\pi \: rdr) {r}^{2}  \\  \\ dI = 2\pi \rho {r}^{3} dr

Integrate both sides

  \int dI =  \int _{0} ^{R}  2\pi \rho {r}^{3} dr\\  \\  I = 2 \pi \rho. \frac{ {R}^{4} }{4}  =  \frac{ \pi \rho R^{4}}{2}

Put value of density from first expression

I =  \frac{\pi {R}^{4} }{2} . \frac{M}{ \pi \: R { }^{2} }  \\  \\ I =  \frac{MR {}^{2} }{2}

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