Spools and Advanced Rotation Problems | JEE Notes

Physical pendulum: where a rigid body behaves like an oscillator

A rigid body swings about a fixed horizontal axis. Its centre of mass is distance $d > 0$ below the pivot at stable equilibrium. Gravity supplies restoring torque $- M g d \sin θ$ :

I_{O} \ddot{θ} = - M g d \sin θ .

For small angular amplitude in radians, $\sin θ \approx θ$ , giving simple harmonic motion:

T = 2 π \sqrt{\frac{I_{O}}{M g d}}, ℓ_{e q u i v a l e n t} = \frac{I_{O}}{M d} .

If $I_{C} = M k^{2}$ , the parallel-axis theorem gives $ℓ_{e q u i v a l e n t} = d + k^{2} / d$ . The period is smallest at $d = k$ , provided such a pivot position is physically available. Large-amplitude oscillations are not exactly SHM; the displayed period is a small-amplitude approximation.

Example 25 · Where should a rod be pivoted to swing fastest?

A uniform slender rod of length $L$ can be pivoted anywhere along its length. Determine the pivot location giving the minimum small-oscillation period, and compare this period with end suspension.

Hint

Minimize the equivalent length $d + k^{2} / d$ , rather than differentiating the full square-root expression for the period.

Complete solution

Since $I_{C} = M L^{2} / 12$ , $k = L / \sqrt{12}$ . By $d + k^{2} / d \geq 2 k$ , equality occurs at $d = k = L / \sqrt{12}$ from the centre. This point lies on the rod because $L / \sqrt{12} < L / 2$ . Orient the rod so its centre hangs below the pivot.

The minimum equivalent length is $2 k = L / \sqrt{3}$ , so $T_{\min} = 2 π \sqrt{L / (\sqrt{3} g)}$ . End suspension has $I_{O} = M L^{2} / 3$ , $d = L / 2$ , and $T_{e n d} = 2 π \sqrt{2 L / (3 g)}$ . Therefore $T_{\min} / T_{e n d} = \sqrt{\sqrt{3} / 2} \approx 0.931$ .

The spool paradox: specify how the string leaves the hub

A spool has outer rolling radius $R$ , inner hub radius $r < R$ , and $I_{C} = β M R^{2}$ . A string leaves the lower side of the inner hub and is pulled with force $F$ at angle $θ$ above the rightward horizontal, where $0 \leq θ < 90^{\circ}$ . Its torque about the centre is anticlockwise, magnitude $F r$ . Changing the winding side changes the problem.

Watch the animation: The spool paradox

Try the simulator: The spool

The force's line of action is tangent to the inner hub. The friction arrow shows the direction obtained for this specified geometry.

Take rightward translation and clockwise rotation positive, retaining signed rightward friction $f$ . While rolling,

F \cos θ + f = M a, - F r - R f = I_{C} (a / R),

a = \frac{F (\cos θ - r / R)}{M (1 + β)}, f = - \frac{F (β \cos θ + r / R)}{1 + β} .

The normal is $N = M g - F \sin θ$ . Require $N > 0$ and $| f | \leq μ_{s} N$ . From rest, the spool accelerates right if $\cos θ > r / R$ , left if $\cos θ < r / R$ , and remains at rest at $θ_{c} = \cos^{- 1} (r / R)$ if friction is sufficient.

At the critical angle, the force line passes through the ground contact. Its torque about that contact vanishes. Friction is nevertheless generally nonzero: it cancels the horizontal force while also balancing the torque about the centre. Zero acceleration is not evidence of zero forces.

Example 26 · Pull right; watch the spool accelerate left

A spool has $M = 2 k g$ , $β = 1 / 2$ , $r / R = 0.60$ , and is pulled in the specified lower-hub geometry by $F = 6 N$ at $60^{\circ}$ . With $g = 10 m s^{- 2}$ and $μ_{s} = 0.30$ , find the acceleration and verify rolling.

Hint

Determine the sign from $\cos θ - r / R$ , then check the reduced normal force.

Complete solution

$a = 6 (0.50 - 0.60) / [2 (1.5)] = - 0.20 m s^{- 2}$ : leftward. Friction is $f = - 6 [(0.5) (0.5) + 0.60] / 1.5 = - 3.40 N$ . The normal force is $N = 20 - 6 \sin 60^{\circ} = 20 - 3 \sqrt{3} \approx 14.80 N$ . Available static friction is $0.30 N \approx 4.44 N > 3.40 N$ , so the rolling solution is valid.

The critical angle is $\cos^{- 1} (0.60) \approx {53.13}^{\circ}$ , below the applied angle. Translation check: $F_{x} + f = 3.0 - 3.4 = - 0.4 N = M a$ .

Optional enrichment · A rolling vertical loop has a hidden friction condition

Consider an ideal circular body rolling inside a rigid vertical circular track of inner radius $A$ . Its centre follows radius $ρ = A - R$ , not $A$ . Let $θ$ be measured from the bottom and let its total mechanical energy equal $M g h$ , with gravitational potential zero at the bottom position of the centre. Assume no dissipative loss and continuous pure rolling with a radial normal.

\frac{v^{2}}{g ρ} = \frac{2 [h / ρ - 1 + \cos θ]}{1 + β}, \frac{N}{M g} = \frac{v^{2}}{g ρ} + \cos θ .

The usual top-contact test $v_{t o p}^{2} \geq g ρ$ gives $h \geq (5 + β) ρ / 2$ . This is the ideal energetic threshold. It does not by itself guarantee that finite static friction can maintain rolling everywhere.

The required tangential friction has magnitude $| f | = β M g | \sin θ | / (1 + β)$ . Combining $| f | \leq μ_{s} N$ over the complete loop gives the stronger finite-friction condition

\frac{h}{ρ} \geq 1 + \frac{12}{\sqrt{(3 + β)^{2} + (β / μ_{s})^{2}}} (μ_{s} > 0) .

To see this, multiply the friction inequality by $1 + β$ : $μ_{s} [2 h / ρ - 2 + (3 + β) \cos θ] \geq β | \sin θ |$ . The largest value of $β | \sin θ | - μ_{s} (3 + β) \cos θ$ is $\sqrt{β^{2} + μ_{s}^{2} (3 + β)^{2}}$ . At any finite $μ_{s}$ and $β > 0$ , this bound exceeds the contact-only threshold. Near the top at the contact-only threshold, required friction falls linearly with angular distance from the top while the normal falls quadratically.

Model boundary: this result concerns the circular section under the assumptions stated. An approach ramp or transition can impose additional friction requirements. Treat the familiar minimum-height answer as conditional on the problem's assumed rolling model.

Rolling with a spring

A body rolling on a floor with its axle on a spring has energy $\frac{12}{(} 1 + β) M {\dot{x}}^{2} + \frac{12}{k} x^{2}$ . The rolling term acts like a heavier mass, so

ω = \sqrt{\frac{k}{(1 + β) M}} .

A solid cylinder gives

ω = \sqrt{2 k / 3 M}

Torsional pendulum

A body on a wire has $τ = - κ θ$ , so $ω = \sqrt{κ / I}$ . For two masses on a light rod at their centre of mass, $I = μ L^{2}$ .

Build the unfamiliar problem from familiar equations