Question:
A gear train with five gears is shown in the figure below.
The number of teeth on each gear is \(N_1\), \(N_2\), \(N_3\), \(N_4\), and \(N_5\). The idler gear in this gear train is
A gear train with five gears is shown in the figure below.
The number of teeth on each gear is \(N_1\), \(N_2\), \(N_3\), \(N_4\), and \(N_5\). The idler gear in this gear train is
The number of teeth on each gear is \(N_1\), \(N_2\), \(N_3\), \(N_4\), and \(N_5\). The idler gear in this gear train is
Show Hint
Write the overall speed ratio from gear 1 to gear 5 and see which gear's tooth count cancels out.
Updated On: Jul 27, 2026
- gear 4
- gear 2
- gear 3
- gear 1
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The Correct Option is A
Solution and Explanation
Step 1: Identify the compound gear and the mesh order.
Gear 1 is the driver and meshes with gear 2. Gears 2 and 3 sit on the same shaft, so they turn together as a compound gear, \(\omega_2 = \omega_3\). Gear 3 meshes with gear 4, and gear 4 meshes with gear 5, the driven gear.
Step 2: Write the speed relation at each mesh.
At a simple external mesh the speed ratio is inverse to the tooth ratio: \(\dfrac{\omega_1}{\omega_2} = \dfrac{N_2}{N_1}\), \(\omega_2 = \omega_3\), \(\dfrac{\omega_3}{\omega_4} = \dfrac{N_4}{N_3}\), \(\dfrac{\omega_4}{\omega_5} = \dfrac{N_5}{N_4}\).
Step 3: Combine the ratios and see what cancels.
Multiplying all four relations gives \(\dfrac{\omega_1}{\omega_5} = \dfrac{N_2}{N_1}\times\dfrac{N_4}{N_3}\times\dfrac{N_5}{N_4} = \dfrac{N_2 N_5}{N_1 N_3}\). The term \(N_4\) drops out completely, so resizing gear 4 changes only its own speed, never the overall ratio between gear 1 and gear 5. A gear whose tooth count cancels like this is called an idler gear.
Final Answer:
Since \(N_4\) plays no role in the input to output speed ratio, gear 4 is the idler. \[ \boxed{\text{Idler gear} = \text{gear } 4} \]
Gear 1 is the driver and meshes with gear 2. Gears 2 and 3 sit on the same shaft, so they turn together as a compound gear, \(\omega_2 = \omega_3\). Gear 3 meshes with gear 4, and gear 4 meshes with gear 5, the driven gear.
Step 2: Write the speed relation at each mesh.
At a simple external mesh the speed ratio is inverse to the tooth ratio: \(\dfrac{\omega_1}{\omega_2} = \dfrac{N_2}{N_1}\), \(\omega_2 = \omega_3\), \(\dfrac{\omega_3}{\omega_4} = \dfrac{N_4}{N_3}\), \(\dfrac{\omega_4}{\omega_5} = \dfrac{N_5}{N_4}\).
Step 3: Combine the ratios and see what cancels.
Multiplying all four relations gives \(\dfrac{\omega_1}{\omega_5} = \dfrac{N_2}{N_1}\times\dfrac{N_4}{N_3}\times\dfrac{N_5}{N_4} = \dfrac{N_2 N_5}{N_1 N_3}\). The term \(N_4\) drops out completely, so resizing gear 4 changes only its own speed, never the overall ratio between gear 1 and gear 5. A gear whose tooth count cancels like this is called an idler gear.
Final Answer:
Since \(N_4\) plays no role in the input to output speed ratio, gear 4 is the idler. \[ \boxed{\text{Idler gear} = \text{gear } 4} \]
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