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.

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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?

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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!