two circular plats of radius R are brought into contact (one on top of the other). The top plate rotates with angular velocity, w, and the bottom plate is fixed. The pressure between the two plates is uniform and constant with a value of p0 and the friction coefficient between the two plates is mu. determine (and show steps) a) the torque acting on the top plate due to the bottom plate b) the power required to keep the top plate spinning at a constant angular velocity, w

Let's break down the problem into parts to find the required quantities: the torque acting on the top plate due to the bottom plate, and the power required to keep the top plate spinning at a constant angular velocity.

### **Given:**
- Radius of each circular plate: \( R \)
- Angular velocity of the top plate: \( \omega \)
- Pressure between the two plates: \( p_0 \) (uniform and constant)
- Coefficient of friction between the plates: \( \mu \)

### **a) Torque Acting on the Top Plate**

The frictional force between the two plates at any point will resist the motion of the top plate. The torque is the result of this frictional force distributed across the surface of the plate.

1. **Frictional Force**:
   At any point on the plate, the frictional force \( f \) is given by:
   \[
   f = \mu \cdot p_0 \cdot dA
   \]
   where \( dA \) is the differential area element.

2. **Torque Contribution**:
   The torque \( d\tau \) due to this frictional force at a distance \( r \) from the center is:
   \[
   d\tau = r \cdot f = r \cdot \mu \cdot p_0 \cdot dA
   \]

3. **Differential Area Element**:
   For a circular plate, the differential area element \( dA \) at a distance \( r \) from the center is:
   \[
   dA = 2\pi r \, dr
   \]

4. **Total Torque**:
   The total torque \( \tau \) acting on the top plate is obtained by integrating the differential torque over the entire surface of the plate:
   \[
   \tau = \int_0^R r \cdot \mu \cdot p_0 \cdot 2\pi r \, dr
   \]
   Simplifying:
   \[
   \tau = 2\pi \mu p_0 \int_0^R r^2 \, dr
   \]
   Evaluating the integral:
   \[
   \tau = 2\pi \mu p_0 \left[ \frac{r^3}{3} \right]_0^R = 2\pi \mu p_0 \cdot \frac{R^3}{3}
   \]
   Thus, the torque acting on the top plate is:
   \[
   \boxed{\tau = \frac{2\pi \mu p_0 R^3}{3}}
   \]

### **b) Power Required to Keep the Top Plate Spinning**

The power required to keep the top plate spinning at a constant angular velocity \( \omega \) is the product of the torque \( \tau \) and the angular velocity \( \omega \):

\[
P = \tau \cdot \omega
\]

Substituting the value of \( \tau \) from part (a):

\[
P = \frac{2\pi \mu p_0 R^3}{3} \cdot \omega
\]

Thus, the power required is:

\[
\boxed{P = \frac{2\pi \mu p_0 R^3 \omega}{3}}
\]

### Summary:
- **Torque acting on the top plate**: \( \tau = \frac{2\pi \mu p_0 R^3}{3} \)
- **Power required to keep the top plate spinning**: \( P = \frac{2\pi \mu p_0 R^3 \omega}{3} \)

---

Would you like to explore more details or have any questions about the derivation? 

Here are five questions you could consider next:
1. How does the torque change if the pressure distribution is not uniform?
2. What happens to the power required if the friction coefficient decreases?
3. How would the analysis change if the bottom plate is not fixed but rotates in the opposite direction?
4. What if the top plate has a non-uniform thickness or mass distribution?
5. Can you derive the expressions for torque and power if the plates were not perfectly circular?

**Tip:** Remember, the torque depends on the distribution of frictional forces over the entire surface of contact. Non-uniform pressure would require a more complex integration.

Learn how to calculate the torque acting on the top plate and the power required to maintain constant angular velocity in a problem involving two rotating circular plates with uniform pressure and friction coefficient. Detailed step-by-step solution provided.

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