Question

A belt drive system given in the figure. Such that distance between the centres $OO'=120cm$ & radii are 20 cm & 80 cm respectively.

Answer 1

3

Answer 2

Solution:

1. Let the given radii be $r = 20$ cm and $R = 80$ cm with center distance $d = 120$ cm.

2. For an external belt, the angle (in each circle) associated with the tangency is given by

   $$\sin\theta = \frac{R - r}{d} = \frac{80 - 20}{120} = \frac{60}{120} = \frac{1}{2}\quad \Rightarrow\quad \theta = \frac{\pi}{6}.$$

3. The length of each straight (tangent) segment is

   $$L_{\text{tangent}} = \sqrt{d^2 - (R - r)^2} = \sqrt{120^2 - 60^2} = \sqrt{14400 - 3600} = 60\sqrt{3}\text{ cm}.$$

4. The belt touches the smaller circle over an arc subtending an angle

   $$\theta_{\text{small}} = 2\pi - 2\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}.$$

   Its arc length is

   $$L_{\text{small}} = r \cdot \theta_{\text{small}} = 20 \times \frac{5\pi}{3} = \frac{100\pi}{3}.$$

5. The belt touches the larger circle over an arc subtending an angle

   $$\theta_{\text{large}} = 2\theta = \frac{\pi}{3}.$$

   Its arc length is

   $$L_{\text{large}} = R \cdot \theta_{\text{large}} = 80 \times \frac{\pi}{3} = \frac{80\pi}{3}.$$

6. The total belt length is the sum of the two arcs and the two tangent segments:

   $$\alpha = L_{\text{small}} + L_{\text{large}} + 2L_{\text{tangent}} = \frac{100\pi}{3} + \frac{80\pi}{3} + 2(60\sqrt{3}) = \frac{180\pi}{3} + 120\sqrt{3} = 60\pi + 120\sqrt{3}.$$

7. Now, calculate $\frac{\alpha}{120}$:

   $$\frac{\alpha}{120} = \frac{60\pi}{120} + \frac{120\sqrt{3}}{120} = \frac{\pi}{2} + \sqrt{3}.$$

   Numerically,

   $$\frac{\pi}{2} \approx 1.5708,\quad \sqrt{3} \approx 1.732,\quad \text{thus} \quad \frac{\alpha}{120} \approx 3.3028.$$

8. Taking the greatest integer function:

   $$\left[\frac{\alpha}{120}\right] = 3.$$

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Core Explanation (Minimal):

• Compute $\theta=\frac{\pi}{6}$ from $\sin\theta=\frac{60}{120}$.
• Tangent segment length: $60\sqrt{3}$ cm.
• Arc on small circle: $20\cdot\frac{5\pi}{3}=\frac{100\pi}{3}$; on large circle: $80\cdot\frac{\pi}{3}=\frac{80\pi}{3}$.
• Total belt length: $60\pi+120\sqrt{3}$ cm.
• Then, $\frac{\alpha}{120}=\frac{\pi}{2}+\sqrt{3}\approx3.3028$ so $\left[\frac{\alpha}{120}\right]=3$.