Source: Final Exam Formula Booklet
Tags: moment of inertia, rotational motion, torque, angular momentum, parallel axis theorem, cross product, rotational kinetic energy, ENGR 216, PHYS 216, Texas A&M
Rotational dynamics is the angular equivalent of Newton's laws for linear motion. The moment of inertia plays the same role as mass, torque replaces force, and angular momentum replaces linear momentum. Knowing which moment-of-inertia formula to pick for a given shape is half the battle on exam problems.
Moment of inertia (I)
A measure of how much an object resists angular acceleration about a given axis. Depends on both the mass distribution and the choice of axis. Units: kg m².
Torque (τ)
The rotational equivalent of force. Produces angular acceleration. Calculated as the cross product of the position vector and the force vector: τ = r × F.
Angular momentum (L)
The rotational equivalent of linear momentum. For a rigid body spinning about a fixed axis: L = Iω.
Parallel axis theorem
A shortcut for finding the moment of inertia about any axis parallel to one through the centre of mass: I_p = I_cm + Md².
Angular velocity (ω)
The rate of change of angular position. Related to linear velocity by v = rω.
Angular acceleration (α)
The rate of change of angular velocity. Related to torque by τ = Iα.
Each formula below gives I about the axis shown on the exam formula sheet. The axis matters: the same object has different I values about different axes.
Thin cylindrical shell (axis along the cylinder's length)
I = MR²
Solid cylinder (axis along the cylinder's length)
I = (1/2)MR²
Solid cylinder (axis perpendicular to length, through centre)
I = (1/4)MR² + (1/12)ML²
Thick cylindrical shell / annular cylinder (axis along length)
I = (1/2)M(R₁² + R₂²)
R₁ is inner radius, R₂ is outer radius
Thick cylinder (axis perpendicular to length, through centre)
I = (1/4)M(R₁² + R₂²) + (1/12)ML²
Hollow sphere (axis through centre)
I = (2/3)MR²
Solid sphere (axis through centre)
I = (2/5)MR²
Thin rod (axis perpendicular to rod, through centre)
I = (1/12)ML²
Thin rod (axis perpendicular to rod, through one end)
I = (1/3)ML²
Rectangular plate (axis through centre, perpendicular to plate)
I = (1/12)M(a² + b²)
a and b are the side lengths
Rectangular plate (axis along one edge)
I = (1/3)Ma²
Point mass at distance R from axis
I = MR²
Disk
A disk is a cylinder of negligible length. Set L = 0 in the cylinder formulae.
For a collection of point masses m₁, m₂, ..., mᵢ at distances r₁, r₂, ..., rᵢ from the axis:
I = Σ mᵢrᵢ²
I_p = I_cm + Md²
I_cm is the moment of inertia about an axis through the centre of mass
d is the perpendicular distance between the two parallel axes
Use this when the rotation axis does not pass through the centre of mass
Vector form: τ = r × F
Magnitude: τ = rF sin φ, where φ is the angle between r and F
Newton's second law for rotation: τ = Iα
Vector form: L = r × p = r × mv
Magnitude: L = mvr sin φ
For a rigid body about a fixed axis: L = Iω
KE_rot = (1/2)Iω²
Often combined with translational KE in rolling-without-slipping problems: KE_total = (1/2)mv² + (1/2)Iω²
a × b = (a_y b_z − a_z b_y) i − (a_x b_z − a_z b_x) j + (a_x b_y − a_y b_x) k
The result is a vector perpendicular to both a and b
Order matters: a × b = −(b × a)
1 revolution = 2π radians
ω = 2πf = 2π/T
v = rω, s = rθ
All moment-of-inertia formulas are tabulated on the exam formula sheet with diagrams showing the rotation axis for each shape. The key distinction to watch: "axis along length" vs "axis perpendicular to length, through centre" gives very different results for cylinders and rods.
⚠️ The most common mistake is using the wrong moment-of-inertia formula because the rotation axis was misidentified. Always check which axis the problem specifies.
⚠️ A disk is just a cylinder with L = 0. Set L = 0 in the cylinder formula rather than hunting for a separate "disk" entry.
⚠️ The parallel axis theorem only works when one of the two axes passes through the centre of mass. You cannot use it to go from one arbitrary axis to another in a single step.
⚠️ For torque and angular momentum magnitudes, sin φ is the angle between the two vectors, not the angle to the horizontal or vertical.
⚠️ Cross products are not commutative. Swapping the order flips the sign.
Q: A solid sphere of mass 4 kg and radius 0.3 m rotates about an axis through its centre. What is its moment of inertia?
A: I = (2/5)(4)(0.3²) = (2/5)(4)(0.09) = 0.144 kg m²
Q: A thin rod of mass 2 kg and length 1.5 m rotates about an axis through its centre. What is I? What if the axis is through one end instead?
A: Through centre: I = (1/12)(2)(1.5²) = 0.375 kg m². Through one end: I = (1/3)(2)(1.5²) = 1.5 kg m².
Q: You know I_cm for an object. The object rotates about a parallel axis 0.5 m away. Mass is 3 kg and I_cm = 0.2 kg m². What is I about the new axis?
A: I_p = 0.2 + 3(0.5²) = 0.2 + 0.75 = 0.95 kg m²
Q: A force of 10 N is applied at a distance of 0.4 m from the pivot, at an angle of 30° to the position vector. What is the torque magnitude?
A: τ = rF sin φ = (0.4)(10) sin 30° = (0.4)(10)(0.5) = 2.0 N m
Q: Is a × b the same as b × a?
A: No. a × b = −(b × a). The direction reverses.
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