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A simple pendulum is set - up in a trolley which moves to the right with an acceleration 'a' on a horizontal plane, then the thread of the pendulum in the mean position makes an angle θ with the vertical
- a)tan-1 (a/g) in the forward direction
- b)tan-1 (a/g) in the backward direction
- c)tan-1 (g/a) in the backward direction
- d)tan-1 (g/a) in the forward direction
Correct answer is option 'B'. Can you explain this answer?
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A simple pendulum is set - up in a trolley which moves to the right wi...
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A simple pendulum is set - up in a trolley which moves to the right wi...
With the vertical. Let's call this angle θ.
When the trolley is at rest (a=0), the pendulum hangs vertically downwards and θ=0.
When the trolley is moving with acceleration a, the pendulum is no longer vertical and θ is non-zero.
To analyze the motion of the pendulum in this situation, we need to consider the forces acting on it. There are two main forces: the tension in the thread and the gravitational force.
The tension in the thread can be decomposed into two components: one along the thread (Tcosθ) and one perpendicular to the thread (Tsinθ). The gravitational force acts downwards with a magnitude of mg, where m is the mass of the pendulum bob and g is the acceleration due to gravity.
In the mean position, the net force acting on the pendulum bob is the sum of the tension component along the thread and the gravitational force. Since the bob is not moving vertically, the net force acting on it must be zero.
Therefore, we can write the following equation:
Tcosθ - mg = 0
This equation tells us that the tension component along the thread is equal to the weight of the pendulum bob (mg). From this equation, we can solve for T:
T = mg/cosθ
Now, let's consider the forces acting along the horizontal direction. The only force in this direction is the tension component perpendicular to the thread (Tsinθ). This force is responsible for providing the necessary centripetal force to keep the bob moving in a circular path.
The centripetal force is given by the equation:
Tsinθ = m * (a - g)
This equation tells us that the tension component perpendicular to the thread is equal to the mass of the bob times the difference between the acceleration of the trolley and the acceleration due to gravity (a - g).
Combining these two equations, we can eliminate T and solve for θ:
mg/cosθ * sinθ = m * (a - g)
Simplifying this equation, we get:
tanθ = (a - g)/g
This equation relates the angle θ with the acceleration of the trolley and the acceleration due to gravity. By knowing the values of a and g, we can determine the angle θ.
When the trolley is at rest (a=0), the pendulum hangs vertically downwards and θ=0.
When the trolley is moving with acceleration a, the pendulum is no longer vertical and θ is non-zero.
To analyze the motion of the pendulum in this situation, we need to consider the forces acting on it. There are two main forces: the tension in the thread and the gravitational force.
The tension in the thread can be decomposed into two components: one along the thread (Tcosθ) and one perpendicular to the thread (Tsinθ). The gravitational force acts downwards with a magnitude of mg, where m is the mass of the pendulum bob and g is the acceleration due to gravity.
In the mean position, the net force acting on the pendulum bob is the sum of the tension component along the thread and the gravitational force. Since the bob is not moving vertically, the net force acting on it must be zero.
Therefore, we can write the following equation:
Tcosθ - mg = 0
This equation tells us that the tension component along the thread is equal to the weight of the pendulum bob (mg). From this equation, we can solve for T:
T = mg/cosθ
Now, let's consider the forces acting along the horizontal direction. The only force in this direction is the tension component perpendicular to the thread (Tsinθ). This force is responsible for providing the necessary centripetal force to keep the bob moving in a circular path.
The centripetal force is given by the equation:
Tsinθ = m * (a - g)
This equation tells us that the tension component perpendicular to the thread is equal to the mass of the bob times the difference between the acceleration of the trolley and the acceleration due to gravity (a - g).
Combining these two equations, we can eliminate T and solve for θ:
mg/cosθ * sinθ = m * (a - g)
Simplifying this equation, we get:
tanθ = (a - g)/g
This equation relates the angle θ with the acceleration of the trolley and the acceleration due to gravity. By knowing the values of a and g, we can determine the angle θ.
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A simple pendulum is set - up in a trolley which moves to the right with an acceleration 'a' on a horizontal plane, then the thread of the pendulum in the mean position makes an angle θ with the verticala)tan-1(a/g)in the forward directionb)tan-1(a/g)in the backward directionc)tan-1(g/a)in the backward directiond)tan-1(g/a)in the forward directionCorrect answer is option 'B'. Can you explain this answer?
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