Four linked discs

Problem

Four discs are connected as follows:

• Disc 1 (radius $r_1 = 1\;\text{m}$) and disc 2 (radius $r_2 = 0{,}5\;\text{m}$) are in edge contact (gears).
• Disc 1 and disc 3 (radius $r_3 = 0{,}3\;\text{m}$) are coaxial (mounted on the same shaft, rigidly attached).
• Disc 3 and disc 4 (radius $r_4 = 0{,}1\;\text{m}$) are connected by a belt pulley.

Disc 2 rotates with $\omega_2 = -2\;\text{rad/s}$ (clockwise). Find the angular velocities of the other three discs.

The three types of connection: gears (1–2), coaxial (1–3), belt pulley (3–4).

Solution

Rules for the three connections. Gears (edge contact): same tangential speed $v$, opposite signs of $\omega$. Coaxial: same $\omega$ (same magnitude and sign), different $v$. Belt pulley: same tangential speed $v$, same sign of $\omega$.

Step 1 — from disc 2 to disc 1 (gears). Same $v$ at the contact point: $\abs{v_1} = \abs{v_2} = \abs{\omega_2}\,r_2 = 2 \cdot 0{,}5 = 1\;\text{m/s}$ $\abs{\omega_1} = \frac{\abs{v_1}}{r_1} = \frac{1}{1} = 1\;\text{rad/s}$ Gears turn in opposite directions, so $\omega_1 = +1\;\text{rad/s}$ (anticlockwise).

Step 2 — from disc 1 to disc 3 (coaxial). Same angular velocity: $\omega_3 = \omega_1 = +1\;\text{rad/s}$ $v_3 = \omega_3\,r_3 = 1 \cdot 0{,}3 = 0{,}3\;\text{m/s}$ (the tangential speed of the rim of disc 3 differs from $v_1$, because the radius is different).

Step 3 — from disc 3 to disc 4 (belt pulley). Same belt speed $v$: $\abs{v_4} = \abs{v_3} = 0{,}3\;\text{m/s}$ $\abs{\omega_4} = \frac{\abs{v_4}}{r_4} = \frac{0{,}3}{0{,}1} = 3\;\text{rad/s}$ The belt transmits the same direction: $\omega_4 = +3\;\text{rad/s}$ (anticlockwise).

Summary.

	Disc 1	Disc 2	Disc 3	Disc 4
$r$ (m)	$1$	$0{,}5$	$0{,}3$	$0{,}1$
$\omega$ (rad/s)	$+1$	$-2$	$+1$	$+3$
$v$ (m/s)	$1$	$1$	$0{,}3$	$0{,}3$

Checking the rules.

• Discs 1–2 (gears): $\abs{v_1} = \abs{v_2} = 1$ ✓; $\omega_1 = +1$, $\omega_2 = -2$, opposite signs ✓.
• Discs 1–3 (coaxial): $\omega_1 = \omega_3 = +1$ ✓; $v_1 = 1$, $v_3 = 0{,}3$, different ✓.
• Discs 3–4 (belt pulley): $\abs{v_3} = \abs{v_4} = 0{,}3$ ✓; $\omega_3 = +1$, $\omega_4 = +3$, same sign ✓.

Result. $\ev{\omega_1 = +1\;\text{rad/s}, \quad \omega_3 = +1\;\text{rad/s}, \quad \omega_4 = +3\;\text{rad/s}}$

Observation: disc 4 is the fastest in $\omega$ despite having the smallest radius. This is the principle of mechanical transmission — transferring motion from one angular velocity to another using pairs of discs (like the gear ratios on a bicycle).

Linked items

Topics: Dinamica Concepts: Moto circolare uniforme · Velocità angolare Skills: Impostazione simbolica Objects: Ingranaggi e dischi coassiali · Disco rotante · Puleggia