Question

An engine running at 150 rpm drives a line shaft by means of a belt. The engine pulley is 750 mm diameter and the pulley on the line shaft being 450 mm. A 900 mm diameter pulley on the line shaft drives a 150 mm pulley keyed to a dynamo shaft. The speed of dynamo shaft in rpm will be

• A. 750
• B. 1000
• C. 1500
• D. 3000

Hint:

For compound belt drives, you can find the final speed in a single step using the compound ratio:
\[ N_{\text{final}} = N_{\text{initial}} \times \left( \frac{d_1}{d_2} \times \frac{d_3}{d_4} \right) \] \[ N_{\text{final}} = 150 \times \left( \frac{750}{450} \times \frac{900}{150} \right) = 150 \times \left( \frac{5}{3} \times 6 \right) = 150 \times 10 = 1500 \text{ rpm} \]

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem represents a compound belt drive system.
A compound drive consists of multiple pulleys where some intermediate shafts carry more than one pulley rotating at the same speed.
Key Formula or Approach:
For any belt drive connecting two pulleys, the speed ratio is inversely proportional to their diameters:
\[ \frac{N_{\text{driven}}}{N_{\text{driver}}} = \frac{d_{\text{driver}}}{d_{\text{driven}}} \] For a compound system:
\[ \frac{N_{\text{final}}}{N_{\text{initial}}} = \frac{\text{Product of diameters of all drivers}}{\text{Product of diameters of all driven pulleys}} \]

Step 2: Detailed Explanation:
Let us identify the components of the drive train:
1. First Stage (Engine to Line Shaft):
- Driver 1 (Engine Pulley): \(d_1 = 750 \text{ mm}\), \(N_1 = 150 \text{ rpm}\).
- Driven 1 (Line Shaft Pulley): \(d_2 = 450 \text{ mm}\), \(N_2\) is the speed of the line shaft.
Using the speed ratio formula:
\[ \frac{N_2}{N_1} = \frac{d_1}{d_2} \implies N_2 = N_1 \times \frac{d_1}{d_2} \] \[ N_2 = 150 \times \frac{750}{450} = 150 \times \frac{5}{3} = 250 \text{ rpm} \] 2. Second Stage (Line Shaft to Dynamo Shaft):
- The second pulley is keyed to the same line shaft, so its speed is the same:
\[ N_3 = N_2 = 250 \text{ rpm} \] - Driver 2 (Line Shaft Pulley): \(d_3 = 900 \text{ mm}\).
- Driven 2 (Dynamo Pulley): \(d_4 = 150 \text{ mm}\), \(N_4\) is the speed of the dynamo.
Using the speed ratio formula:
\[ \frac{N_4}{N_3} = \frac{d_3}{d_4} \implies N_4 = N_3 \times \frac{d_3}{d_4} \] \[ N_4 = 250 \times \frac{900}{150} = 250 \times 6 = 1500 \text{ rpm} \] The rotational speed of the dynamo shaft is 1500 rpm.

Step 3: Final Answer:
The correct option is (C).