Question

A pair of straight teeth spur gears is transmitting power at 600 rpm and the pinion has 21 standard full depth involute teeth of module 7 mm. The pitch line velocity of the pinion is

• A. 0.09 \(m\cdot s^{-1}\)
• B. 0.22 \(m\cdot s^{-1}\)
• C. 2.32 \(m\cdot s^{-1}\)
• D. 4.62 \(m\cdot s^{-1}\)

Hint:

To speed up calculations, simplify the division first:
\(\frac{600}{60} = 10\).
Then multiply by the diameter: \(10 \times 0.147 = 1.47\).
Finally multiply by \(\pi \approx 22/7\): \(1.47 \times \frac{22}{7} = 0.21 \times 22 = 4.62\text{ m/s}\).

Answer 1

The Correct Option is D

Step 1: Recover the pitch circle diameter from the gear data.
For standard involute gears the pitch diameter is module times number of teeth, so \( D = m \times T = 7 \times 21 = 147 \text{mm} = 0.147 \text{m} \).
Step 2: Apply the pitch line velocity relation.
Pitch line velocity is the linear speed of a point on the pitch circle, given by \( v = \dfrac{\pi D N}{60} \), where N is in rpm.
Step 3: Substitute and compute.
\( v = \dfrac{\pi \times 0.147 \times 600}{60} = \pi \times 0.147 \times 10 = 4.618 \text{m/s} \), which rounds to 4.62 m/s.
\[ \boxed{4.62 \ m \cdot s^{-1}} \]