Let's break down the problem and solve it step by step: --- ## **Given Data** | Component | Parameter | Value | |---------------|----------------------------------|---------| | Spur Gear | Pitch diameter (\(d_g\)) | 20 cm | | Spur Gear | Number of teeth (\(n_g\)) | 72 | | Worm | Diameter (\(d_m\)) | 14 mm | | Worm | No. of thread starts (\(n_{pw}\))| 2 | | Gear Interface| Coefficient of friction (\(\mu\))| 0.17 | | | Required table rpm | 5 rpm | | | Friction torque on gear (\(T_f\))| 300 mN·m| --- # **a. Find Worm Shaft RPM for Table at 5 RPM** ### **Step 1: Worm Gear Ratio** The worm and worm gear mechanism has a gear ratio calculated as: \[ \text{Gear Ratio} = \frac{\text{Number of teeth on gear}~(n_g)}{\text{Number of worm threads}~(n_{pw})} \] \[ \text{Gear Ratio} = \frac{72}{2} = 36 \] --- ### **Step 2: Relate Output and Input RPM** For every complete revolution of the worm, the gear advances by the number of threads. So: \[ \text{Output rpm} = \frac{\text{Input rpm}}{\text{Gear Ratio}} \] Let \(N_{worm}\) = rpm of the worm shaft (motor input) Let \(N_{table}\) = rpm of the table (gear output) = 5 rpm \[ N_{table} = \frac{N_{worm}}{\text{Gear Ratio}} \] \[ 5 = \frac{N_{worm}}{36} \] \[ N_{worm} = 5 \times 36 = 180~\text{rpm} \] **Final Answer (a):** > **The worm shaft (motor input) must rotate at 180 rpm for the table to rotate at 5 rpm.** --- # **b. Torque Required at the Worm (Motor Input)** Given: - Friction torque on gear: \(T_f = 300~\text{mN}\cdot\text{m} = 0.3~\text{N}\cdot\text{m}\) - Coefficient of friction (\(\mu\)) = 0.17 - Mean worm diameter (\(d_m\)) = 14 mm = 0.014 m - Number of worm threads (\(n_{pw}\)) = 2 ## **Step 1: Calculate Lead Angle (\(\lambda\))** \[ \text{Lead} = \text{(Pitch)} \times \text{Number of threads} \] Pitch circumference (\( = \pi d_m \)) \[ \text{Lead} = \frac{n_{pw} \times \pi d_m}{n_g} \] But actually, for lead angle: \[ \tan\lambda = \frac{\text{Lead}}{\pi d_m} \] Lead = n_{pw} \times (\text{Pitch}) But since we don't have the pitch, use: \[ \text{Lead} = n_{pw} \times \text{(Pitch)} \] But in gear design, the lead angle (\(\lambda\)) is: \[ \tan\lambda = \frac{n_{pw}}{n_g} \cdot d_g / d_m \] But for simplicity, let's use the form: \[ \tan\lambda = \frac{\text{Lead}}{\pi d_m} \] Lead = n_{pw} \times (\pi d_m / n_g) \times n_g = n_{pw} \times \pi d_m So, \[ \tan\lambda = \frac{n_{pw} \cdot \pi d_m}{\pi d_m} = n_{pw} \] But this gives too high, so let's use the basic form: \[ \text{Lead} = n_{pw} \times \text{Pitch} \] We don't have the pitch, but typically, let's just work with the standard worm gear torque transmission formula: --- ## **Step 2: Torque Transmission for Worm Gear** The input torque required to overcome friction is: \[ T_{in} = T_{out} \cdot \frac{d_m}{2r_g} \cdot \frac{1 + \mu \cot \lambda}{1 - \mu \tan \lambda} \] But a simpler form (assuming no significant efficiency losses other than interface friction): \[ T_{worm} = \frac{T_{gear}}{\eta} \cdot \frac{1}{\text{Gear Ratio}} \] or, with friction, \[ T_{in} = \frac{T_{out}}{\eta} \] where \(\eta\) is the efficiency: \[ \eta = \frac{\tan \lambda}{\tan \lambda + \mu} \] First, find the lead angle (\(\lambda\)): \[ \tan \lambda = \frac{\text{Lead}}{\pi d_m} \] Lead = n_{pw} \times \text{Pitch} Pitch circumference = \(\pi d_m\) Each turn of the worm advances the gear by n_{pw} teeth, so Lead = n_{pw} \times \text{Pitch circumference per tooth} But since pitch per tooth = \(\pi d_g / n_g\): \[ \text{Lead} = n_{pw} \times \frac{\pi d_g}{n_g} \] Plug in values: - \(n_{pw} = 2\) - \(d_g = 0.2~\text{m}\) - \(n_g = 72\) \[ \text{Lead} = 2 \times \frac{\pi \times 0.2}{72} \] \[ \text{Lead} = 2 \times \frac{0.628}{72} = 2 \times 0.00873 = 0.01745~\text{m} \] Now, - \(d_m = 0.014~\text{m}\) \[ \tan \lambda = \frac{0.01745}{\pi \times 0.014} = \frac{0.01745}{0.04398} = 0.397 \] \[ \lambda = \arctan(0.397) = 21.6^\circ \] --- ## **Step 3: Calculate Efficiency (\(\eta\))** \[ \eta = \frac{\tan \lambda}{\tan \lambda + \mu} \] \[ \eta = \frac{0.397}{0.397 + 0.17} = \frac{0.397}{0.567} = 0.700 \] --- ## **Step 4: Input Torque Required at Worm** \[ T_{worm} = \frac{T_{gear}}{\eta} \cdot \frac{1}{\text{Gear Ratio}} \] But the gear ratio is 36. However, the torque multiplication is actually: \[ T_{worm} = \frac{T_{gear}}{\eta \cdot \text{Gear Ratio}} \] But, actually, for a worm gear, the **output torque** is: \[ T_{out} = T_{in} \cdot \eta \cdot \text{Gear Ratio} \] So, \[ T_{in} = \frac{T_{out}}{\eta \cdot \text{Gear Ratio}} \] Plug in: - \(T_{out} = 0.3~\text{N}\cdot\text{m}\) - \(\eta = 0.70\) - \(\text{Gear Ratio} = 36\) \[ T_{in} = \frac{0.3}{0.70 \times 36} \] \[ T_{in} = \frac{0.3}{25.2} = 0.0119~\text{N}\cdot\text{m} \] --- ## **Final Answers** ### **a.** > **The horizontal shaft (worm) must rotate at 180 rpm for the table to rotate at 5 rpm.** ### **b.** > **The torque required at the worm (motor input shaft) is approximately 0.012 N·m (12 mN·m).** --- **Let me know if you want more explanation on any step!**