4. The part of the machine consists of a 50.0 cm long beam, the ends of which are attached with screws
mass is 0.50 kg / piece. The screws can withstand a maximum force of 87.0 N before they come loose. Bar
rotates around an axis passing through its center and perpendicular to the beam.
a) When the beam is rotating at constant speed on a horizontal frictionless surface, what is the maximum speed it can
on pieces without the screws coming off?
b) Suppose the machine is redesigned so that the beam rotates at a constant vertical speed
level. Is the probability of the screw coming off greater when the piece is at the highest point of the track
at or at the lowest point of the track? Use a free-form image and explain why.
c) Use b) to determine the maximum speed the pieces can have without the screw coming loose.

Let's break down each part of the problem to find the solutions. Here’s the approach and solution to each question:

### Given Data
- Length of the beam, \( L = 50.0 \, \text{cm} = 0.50 \, \text{m} \)
- Mass at each end of the beam, \( m = 0.50 \, \text{kg} \)
- Maximum force each screw can withstand, \( F_{\text{max}} = 87.0 \, \text{N} \)
  
Since the beam rotates about an axis passing through its center, the rotation will create a centripetal force that each screw must handle.

### Part (a): Maximum Speed on a Horizontal Surface

When the beam is rotating at constant speed on a horizontal frictionless surface, each mass experiences centripetal force, given by:
\[
F_c = \frac{mv^2}{r}
\]
where:
- \( m \) is the mass at each end,
- \( v \) is the tangential speed, and
- \( r \) is the distance from the center of rotation to each mass (half the length of the beam), \( r = \frac{L}{2} = 0.25 \, \text{m} \).

Since \( F_{\text{max}} = 87.0 \, \text{N} \), we set \( F_c = F_{\text{max}} \) and solve for \( v \):
\[
87.0 = \frac{0.50 \cdot v^2}{0.25}
\]
\[
87.0 = 2.0 \cdot v^2
\]
\[
v^2 = \frac{87.0}{2.0}
\]
\[
v^2 = 43.5
\]
\[
v = \sqrt{43.5} \approx 6.6 \, \text{m/s}
\]
So, the maximum speed the beam can rotate without the screws coming off is approximately **6.6 m/s**.

### Part (b): Vertical Rotation and Probability of Screw Coming Off

When the beam rotates in a vertical plane, the force required to keep each mass in circular motion varies depending on its position along the path:
- At the **lowest point**, the gravitational force adds to the centripetal force, increasing the total force on the screws.
- At the **highest point**, gravity acts opposite to the centripetal force, reducing the total force on the screws.

Thus, the probability of the screws coming loose is higher at the **lowest point of the track**, where the combined force (centripetal force plus gravitational force) is greatest.

### Part (c): Maximum Speed in Vertical Plane

At the lowest point of the vertical path, the total force \( F_{\text{total}} \) on the screws is the sum of the centripetal force and the gravitational force:
\[
F_{\text{total}} = F_c + mg
\]
Setting \( F_{\text{total}} = F_{\text{max}} \), we get:
\[
87.0 = \frac{mv^2}{r} + mg
\]
Substitute \( r = 0.25 \, \text{m} \), \( m = 0.50 \, \text{kg} \), and \( g = 9.8 \, \text{m/s}^2 \):
\[
87.0 = \frac{0.50 \cdot v^2}{0.25} + 0.50 \cdot 9.8
\]
\[
87.0 = 2.0 \cdot v^2 + 4.9
\]
\[
82.1 = 2.0 \cdot v^2
\]
\[
v^2 = \frac{82.1}{2.0}
\]
\[
v^2 = 41.05
\]
\[
v = \sqrt{41.05} \approx 6.4 \, \text{m/s}
\]
Thus, the maximum speed the beam can rotate in a vertical plane without the screws coming loose is approximately **6.4 m/s**.

---

### Summary of Answers:
1. **Maximum speed on a horizontal surface:** \( v \approx 6.6 \, \text{m/s} \)
2. **Greater probability of screws coming loose:** At the lowest point of the vertical rotation.
3. **Maximum speed in a vertical plane:** \( v \approx 6.4 \, \text{m/s} \)

Would you like a detailed explanation of any part of the solution?

---

### Related Questions:
1. How would the solution change if the beam’s mass were included in the calculations?
2. What would happen to the maximum speed if the mass of each end were doubled?
3. How would friction affect the maximum speed on a horizontal surface?
4. What would be the impact on the screws if the rotation axis shifted closer to one end?
5. How would the force calculations change if the screws could only withstand 50 N?

### Tip:
Always separate forces acting in the same and opposite directions when calculating total forces in circular motion problems!

The part of the machine consists of a 50.0 cm long beam, the ends of which are attached with screws mass is 0.50 kg / piece. The screws can withstand a maximum force of 87.0 N before they come loose. Bar rotates around an axis passing through its center and perpendicular to the beam. a) When the beam is rotating at constant speed on a horizontal frictionless surface, what is the maximum speed it can on pieces without the screws coming off? b) Suppose the machine is redesigned so that the beam rotates at a constant vertical speed level. Is the probability of the screw coming off greater when the piece is at the highest point of the track at or at the lowest point of the track? Use a free-form image and explain why. c) Use b) to determine the maximum speed the pieces can have without the screw coming loose.

Solve the problem of calculating the maximum speed of a rotating beam on a horizontal frictionless surface and in a vertical plane. This problem covers concepts like centripetal force, gravity, and forces on screws. Includes a step-by-step solution and explanations of key concepts. Suitable for students in Grades 10-12.

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