Question

moment of inertia of solid cylinder in its own axis derivation and mathematically​

Answer 1

Explanation:

. We will use the general equation of moment of inertia:

dI = r2 dm

Now we move on to finding the dm. It is normally given as;

dm = ρ dV

In this case, the mass element can be expressed in terms of an infinitesimal radial thickness dr by;

dm = 2r L dr

In order to obtain dm we have to calculate dv first. It is given as;

dV = dA L

Meanwhile, dA is the area of the big ring (radius: r + dr) minus the smaller ring (radius: r). Hence;

dA = \pi (r + dr)^{2} – \pi r^{2}dA=π(r+dr)

2

–πr

2

dA = \pi (r + 2rdr + (dr)^{2}) – \pi r^{2}dA=π(r+2rdr+(dr)

2

)–πr

2

Notably, here the (dr)2 = 0.

dA = 2\pi r drdA=2πrdr

Answer 2

Answer:

Explanation:

1. We will use the general equation of moment of inertia:

dI = r² dm

Now we move on to finding the dm. It is normally given as;

dm = ρ dV

In this case, the mass element can be expressed in terms of an infinitesimal radial thickness dr by;

dm = 2r L dr

In order to obtain dm we have to calculate dv first. It is given as;

dV = dA L

Meanwhile, dA is the area of the big ring (radius: r + dr) minus the smaller ring (radius: r). Hence;

dA = π ( r+dr )² - πr²  => dA = π (r + 2r dr + (dr)² ) - πr²

Notably, here the (dr)² = 0.

dA = 2πrdr

2. Substitution of dA into dV we get;

dV = 2πLdr

Now, we substitute dV into dm and we will have;

dm = 2πLdr

The dm expression is further substituted into the dI equation and we get;

dl = 2πr³Ldr

3. Alternatively, we have to find the expression for density as well. We use the equation;

p = M/V

Now,

p = M/πR²L

4. The final step involves using integration to find the moment of inertia of the solid cylinder. The integration basically takes the form of a polynomial integral form.

I = 2PπrL ₙ∫ᵇ r³dr  => I =  2PπL r⁴/4   => I = 2π[m/πR²L] LR⁴/4.

Therefore, I = 1/2 MR²