Q.A uniform rod of mass $m$ and length $\ell$ makes a constant angle $\theta$ with an axis of rotation which passes through one end of the rod. Find the moment of inertia about this axis.

The moment of inertia of a uniform rod of mass M and length l about a perpendicular axis through one end is (1/3)Ml^2, obtained from the centre-of-mass value (1/12)Ml^2 via the parallel axis theorem.

Step 1: Moment of inertia about the centre.

For a uniform rod of mass M and length l, rotating about a perpendicular axis through its geometric centre (centre of mass), the standard result (derived by integrating dI = x^2 dm over the length) is:

I_cm = (1/12) M l^2

Step 2: Apply the parallel axis theorem. …

Moment of inertia of a uniform disc about its central (symmetry) axis.

Treat the disc as a set of concentric thin rings of radius $x$, thickness $dx$, mass $dm=\dfrac{2M}{R^2}x\,dx$ (surface density $\times$ ring area). Each ring's MI about the central axis is $dm\,x^2$. Integrating from $0$ to $R$: …

A disc's moment of inertia about the central perpendicular axis is ½MR², derived by integrating over concentric rings.

Consider a uniform disc of mass M and radius R, lying in a plane, with the axis passing through its centre O and perpendicular to the plane (the axis a student would picture as sticking straight out of the disc's face).

Divide the disc into thin concentric rings of radius x and thickness dx. The mass of this ring is

dm = (M / πR²) × (2πx dx) = (2M/R²) x dx

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MI of a uniform disc about the central axis perpendicular to its plane is (1/2)MR².

For a uniform disc about an axis through its centre, perpendicular to its plane: …

Moment of inertia (I) is the rotational analogue of mass — it measures a rigid body's resistance to a change in its angular velocity, just as mass measures resistance to a change in linear velocity.

In translational (straight-line) motion, Newton's second law is F = ma: mass m is the measure of inertia — how strongly a body resists being accelerated by a force. In rotational motion, the corresponding law is τ = Iα, where torque τ plays the role of force, angular acceleration α plays the role of linear acceleration, and moment of in …

As the rotating cylindrical container's water redistributes outward under the rotation, the moment of inertia of the water-container system increases.

The moment of inertia of a mass distribution about an axis is I = sum(m_i r_i^2), where r_i is each mass element's perpendicular distance from the axis. It depends on HOW mass is distributed relative to the axis, not just the total mass.

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For a thin ring about its central axis (perpendicular to its plane), all the mass lies at the same distance $R$ from the axis, so $I = MR^2$ directly.

For a thin circular ring, every mass element lies at radius $R$ from the axis passing through the centre, perpendicular to the plane of the ring. Hence the moment of inertia is

$I = \sum m_i r_i^2 = MR^2$ …

Parallel axis theorem: I = I_cm + Md².

The theorem of parallel axes states that the moment of inertia of a rigid body about any axis is equal to its moment of inertia about a parallel axis passing through its centre of mass, plus the product of the mass of the body and the square of the perpendicular distance between the two axes.

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Equate rotational KE ($\tfrac12 I\omega^2$) to translational KE ($\tfrac12 mv^2$) and solve for $v$.

Rotational KE of the rotating body:

$KE_{rot}=\frac12 I\omega^2=\frac12\times 5\times 6^2=\frac12\times 5\times 36=90\text{ J}$

This must equal the translational KE of the 20 kg mass: …