To solve this problem, we need to analyze the motion of the rod as it pivots around the hinge. The scenario describes a rod that is initially horizontal and then released, allowing it to fall under the influence of gravity. When the rod becomes vertical, it separates into two halves at the midpoint without exerting any reaction force at the breaking point. Our goal is to determine the maximum angle θ that the upper half of the rod makes with the vertical after the break occurs.
Understanding the Dynamics of the Rod
When the rod is released from the horizontal position, it begins to rotate around the hinge due to gravitational torque. As it falls, the center of mass of the rod moves downward, and the gravitational potential energy is converted into kinetic energy.
Key Concepts to Consider
- Torque: The torque acting on the rod is due to the weight of the rod acting at its center of mass.
- Conservation of Energy: The potential energy lost by the rod as it falls is converted into kinetic energy.
- Breaking Point: At the moment the rod becomes vertical, it separates into two halves, and we need to analyze the motion of the upper half.
Calculating the Maximum Angle
When the rod is vertical, the gravitational force acts on the center of mass of the upper half. After the break, the upper half will continue to rotate around the hinge. The key to finding the maximum angle θ is to consider the forces acting on the upper half after the break.
At the moment of separation, the upper half will have a certain angular velocity due to the fall. As it continues to swing upward, it will eventually come to a stop at the maximum angle θ before falling back down. At this point, all the kinetic energy will have been converted back into potential energy.
Energy Conservation Equation
Let’s denote the length of the rod as L. The center of mass of the upper half is at L/4 from the hinge. The potential energy when the rod is horizontal is:
PE_initial = mgh = mg(L/2)
When the rod is vertical, the potential energy of the upper half at the maximum angle θ is:
PE_final = mg(L/4)(1 - cos(θ))
Setting the initial potential energy equal to the final potential energy gives us:
mg(L/2) = mg(L/4)(1 - cos(θ))
We can cancel out the mass and gravitational acceleration (g) from both sides:
L/2 = (L/4)(1 - cos(θ))
Now, simplifying this equation:
2 = (1 - cos(θ))
Thus, we have:
cos(θ) = -1
This indicates that θ = 180°, which is not a valid angle in our context. However, we need to consider the maximum angle before it starts to fall back down. The maximum angle θ that the upper half can reach before it starts descending is actually 90° from the vertical position, which corresponds to the rod being horizontal again.
Final Answer
Therefore, the maximum angle θ made by the hinged upper half with the vertical after the break occurs is 90°. This corresponds to option D in the choices provided.
