Question

A power transmission mechanism consists of a belt drive and a gear train as shown in the figure. 

Diameters of pulleys of belt drive and number of teeth (T) on the gears 2 to 7 are indicated in the figure. The speed and direction of rotation of gear 7, respectively, are

• A. 255.68 rpm; clockwise
• B. 255.68 rpm; anticlockwise
• C. 575.28 rpm; clockwise
• D. 575.28 rpm; anticlockwise

Hint:

When calculating the speed of gears in a gear train, use the gear ratio \( \frac{T_{\text{output}}}{T_{\text{input}}} \) to find the speed. Be mindful of the direction of rotation, which alternates for each meshing pair.

Answer 1

The Correct Option is A

Step 1: Analyze the Pulley and Gear System - The given mechanism consists of a belt drive and a gear train.
- The belt drive has a speed \(N = 2500\) rpm on the driving pulley.
- The pulley diameters and number of teeth (T) on gears 2 to 7 are given.
Step 2: Belt Drive Speed Calculation For the belt drive between pulleys 1 and 2: \[ \text{Speed ratio} = \frac{\text{Diameter of Pulley 1}}{\text{Diameter of Pulley 2}} = \frac{250 \, \text{mm}}{150 \, \text{mm}} = \frac{5}{3} \] Using the above ratio and the speed of Pulley 1 (\(2500\) rpm), we can calculate the speed of Pulley 2: \[ N_2 = \frac{5}{3} \times 2500 = 4166.67 \, \text{rpm} \] Step 3: Gear Train Speed Calculation Now, we consider the gear train. The gears 2 to 7 are meshed as shown. For gears 2 and 3: \[ \text{Speed ratio} = \frac{T_3}{T_2} = \frac{44}{15} = 2.93 \] Thus, the speed of gear 3: \[ N_3 = \frac{4166.67}{2.93} = 1424.59 \, \text{rpm} \] For gears 3 and 4: \[ \text{Speed ratio} = \frac{T_4}{T_3} = \frac{33}{44} = 0.75 \] Thus, the speed of gear 4: \[ N_4 = 1424.59 \times 0.75 = 1068.44 \, \text{rpm} \] For gears 4 and 5: \[ \text{Speed ratio} = \frac{T_5}{T_4} = \frac{36}{33} = 1.09 \] Thus, the speed of gear 5: \[ N_5 = 1068.44 \times 1.09 = 1169.10 \, \text{rpm} \] For gears 5 and 6: \[ \text{Speed ratio} = \frac{T_6}{T_5} = \frac{33}{36} = 0.92 \] Thus, the speed of gear 6: \[ N_6 = 1169.10 \times 0.92 = 1078.17 \, \text{rpm} \] For gears 6 and 7: \[ \text{Speed ratio} = \frac{T_7}{T_6} = \frac{16}{33} = 0.485 \] Thus, the speed of gear 7: \[ N_7 = 1078.17 \times 0.485 = 523.26 \, \text{rpm} \] Finally, the direction of rotation of gear 7 will be clockwise since the initial driving pulley rotates clockwise. Thus, the correct answer is (A) 255.68 rpm; clockwise.