The Spin Zone: Mastering System of Particles and Rotational Motion

If linear motion was a straight road, Rotational Motion is a mountain pass with hairpin turns. This chapter takes everything you learned about force, mass, and acceleration and “twists” it.

We stop treating objects like tiny dots (point masses) and start treating them like real, extended bodies. This is where we learn why a spinning top doesn’t fall over, why an ice skater spins faster when they pull their arms in, and why a ring loses a race against a solid sphere down a hill.

The Core Pillars of Rotation

1. The Center of Mass (COM)

The COM is the “average” location of the mass in a system. If you throw a spinning wrench through the air, the wrench looks chaotic, but the Center of Mass follows a perfect, calm parabola.

2. Moment of Inertia (I): The Distribution Matters

In linear motion, mass is just mass. In rotation, where that mass is located relative to the axis is everything. This is the Moment of Inertia. A hollow cylinder is harder to start (and stop) spinning than a solid one of the same mass because its mass is further from the center.

3. Torque (τ): The Turning Force

Torque is the rotational equivalent of Force. It’s not just about how hard you push, but where and at what angle.

• τ = r × F = rF sin(θ)This is why doorknobs are on the edge of the door, not near the hinges!

4. Angular Momentum (L) and Conservation

Angular momentum (L = Iω) is the quantity of rotation. If no external torque acts on a system, its angular momentum stays constant. This is the magic behind divers, gymnasts, and even the formation of galaxies.

The Gauntlet: 10 Challenging Aptitude Questions

These questions require you to synthesize multiple concepts (COM, Torque, and Energy).

Question 1: The Missing Piece

A circular disc of radius R has a smaller circular hole of radius R/2 cut out from it. The edge of the hole touches the center of the original disc. Find the shift in the Center of Mass of the remaining portion.

Question 2: The Rod Pivot

A uniform rod of length L and mass M is held horizontally and pivoted at one end. When the rod is released, what is the initial angular acceleration (α)?

Question 3: The Rolling Race

A solid sphere, a solid disc, and a hollow ring—all of the same mass and radius—are released from the top of an incline. In what order will they reach the bottom, and why?

Question 4: The Ice Skater’s Work

An ice skater is spinning with an angular velocity ω with her arms extended. She pulls her arms in, reducing her moment of inertia by half (I/2). By what factor does her Kinetic Energy change? Where does this energy come from?

Question 5: The Unwinding Spool

A spool of mass M and radius R has a string wrapped around it. If you pull the string horizontally with a force F while the spool sits on a frictionless surface, what is the acceleration of the Center of Mass?

Question 6: The Instantaneous Center

A wheel of radius R is rolling without slipping on a horizontal floor with velocity v. What is the velocity of the highest point of the wheel relative to the ground? What about the point in contact with the ground?

Question 7: The Toppling Chimney

A tall vertical chimney of height h falls over from its base (which acts as a pivot). At what height does the chimney experience the greatest tension while falling? (Hint: Consider the chimney as a rigid rod).

Question 8: The Angular Impulse

A square plate of side a and mass m is hit by a sudden impulse J at one of its corners, perpendicular to the side. Describe the resulting motion of the plate’s Center of Mass and its angular velocity.

Question 9: The Yo-Yo Condition

A Yo-Yo of mass M and radius R is allowed to fall. If the string is wound around an inner axle of radius r, find the acceleration of the Yo-Yo as it descends.

Question 10: Radius of Gyration Shift

If the temperature of a solid metal sphere increases, causing its radius to expand by 1%, by what percentage does its Moment of Inertia increase?

Detailed Explanations & Solutions

1. The Missing Piece

Use the formula: x_com = (A₁x₁ – A₂x₂) / (A₁ – A₂).

Let the center of the big disc be (0,0). The hole’s center is at (R/2, 0).

Mass is proportional to Area. Area Ratio is 1 : 1/4.

Shift = [0 – (1/4)(R/2)] / [1 – 1/4] = (-R/8) / (3/4) = -R/6.

Result: The COM moves R/6 away from the hole.

2. The Rod Pivot

Torque τ = Iα.

The weight acts at the COM (L/2). τ = Mg(L/2).

For a rod pivoted at the end, I = (1/3)ML².

Mg(L/2) = (1/3)ML²α → α = 3g / 2L.

3. The Rolling Race

The acceleration of a rolling object is a = g sinθ / (1 + K²/R²), where K is the radius of gyration.

• Sphere: K²/R² = 0.4 (Highest acceleration)
• Disc: K²/R² = 0.5
• Ring: K²/R² = 1.0 (Lowest acceleration)Result: Sphere > Disc > Ring.

4. The Ice Skater

L is conserved (I₁ω₁ = I₂ω₂). If I₂ = I₁/2, then ω₂ = 2ω₁.

KE = ½Iω².

New KE = ½(I/2)(2ω)² = ½I(4ω²)/2 = 2 × (Old KE).

Result: KE doubles. The energy comes from the Internal Work done by the skater’s muscles to pull her arms in.

5. The Unwinding Spool

Since there is no friction, the only horizontal force is F.

F = Ma_com.

Result: a = F/M. (The rotation of the spool doesn’t affect the linear acceleration of the COM if the surface is frictionless).

6. The Instantaneous Center

In pure rolling, the contact point is at rest relative to the ground (v_contact = 0).

The top point has two velocities: translation (v) and rotation (ωR). Since v = ωR,

Result: v_top = 2v; v_bottom = 0.

7. Toppling Chimney

As the chimney falls, different segments have different centripetal requirements.

Result: The chimney usually breaks at about 1/3 of its height from the bottom because the shear stress and bending moment are maximized there.

8. Angular Impulse

Linear: J = M v_com → v_com = J/M.

Angular: Torque × time = ΔL. J × (a/2) = Iω.

For a square plate about its center, I = Ma²/6.

Result: v_com = J/M; ω = 3J / Ma.

9. The Yo-Yo

Equations: Mg – T = Ma and Tr = Iα. Since a = rα:

a = g / (1 + I/Mr²). For a disc-like Yo-Yo, I = ½MR².

Result: a = g / (1 + R²/2r²).

10. Radius of Gyration Shift

I = (2/5)MR².

If R increases by 1%, R_new = 1.01R.

I_new = (2/5)M(1.01R)² ≈ (2/5)M(1.02R²).

Result: 2% increase. (Moment of Inertia is proportional to the square of the radius).

Pro-Tip: The “Translation-Rotation” Dictionary

When you get stuck, translate the problem into linear terms to find the right formula:

• Mass (M) → Moment of Inertia (I)
• Velocity (v) → Angular Velocity (ω)
• Force (F) → Torque (τ)
• Momentum (p) → Angular Momentum (L)