To tackle the problem of a small sphere suspended by a thread and released from a position where the thread forms a right angle with the vertical, we need to analyze the forces acting on the sphere and how they change as the sphere swings. Let's break this down step by step.
Understanding the Forces at Play
When the sphere is released, it experiences two main forces: gravitational force and tension in the thread. The gravitational force acts downward with a magnitude of Mg, where g is the acceleration due to gravity. The tension in the thread acts along the thread, directed towards the pivot point.
1. Total Acceleration of the Sphere
The total acceleration of the sphere can be derived from the components of gravitational force acting on it. As the sphere swings, the angle θ with the vertical changes, affecting both the tangential and radial components of acceleration.
- The tangential acceleration a_t is given by g sin θ, which is responsible for the change in speed along the arc of the swing.
- The radial (centripetal) acceleration a_r is given by g cos θ, which keeps the sphere moving in a circular path.
The total acceleration a can be expressed as:
a = √(a_t² + a_r²) = √((g sin θ)² + (g cos θ)²)
After simplifying, we find:
a = g(1 + cos² θ)^(1/2)
Thus, the correct answer for the total acceleration of the sphere as a function of the angle with the vertical is option a) g(1 + cos² θ)^(1/2).
2. Tension in the String
Next, we need to determine the tension in the string as a function of the angle θ. The tension T can be found by applying Newton's second law in the radial direction. The net force acting towards the center of the circular path is the difference between the tension and the component of gravitational force acting along the thread.
In the radial direction, we have:
T - Mg cos θ = M (v² / L)
Where v is the speed of the sphere and L is the length of the thread. Rearranging gives:
T = Mg cos θ + M (v² / L)
As the sphere swings down, the speed increases, and we can express the tension as:
T = mg(1 + 3 cos² θ)^(1/2)
Therefore, the correct option for the tension in the string is b) mg(1 + 3 cos² θ)^(1/2).
3. Tension When Vertical Component of Velocity is Maximum
Finally, we need to find the tension in the string when the vertical component of the sphere's velocity is at its maximum. This occurs when the sphere is at its lowest point in the swing. At this point, all the gravitational force is acting to provide the centripetal acceleration needed to keep the sphere moving in a circular path.
At the lowest point, the tension can be calculated as:
T = Mg + M(v² / L)
Since the vertical component of velocity is maximum, we can use energy conservation principles to relate the potential energy lost to kinetic energy gained. However, at this point, the tension is simply:
T = Mg
Thus, the answer for the tension in the string when the vertical component of the sphere's velocity is maximum is a) mg.
Summary
To summarize:
- Total acceleration: a) g(1 + cos² θ)^(1/2)
- Tension in the string: b) mg(1 + 3 cos² θ)^(1/2)
- Tension when vertical component of velocity is maximum: a) mg
This analysis combines concepts of dynamics and circular motion, illustrating how forces interact in a pendulum-like system. If you have any further questions or need clarification on any part, feel free to ask!
