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Thin uniform bar of mass m and length 2L is held at an angle son with the horizontal by means of two vertical inextensible strings, at each end as shown in figure. If the string at the right end breaks, leaving the bar to swing, determine the tension in the string at the left end and the angular acceleration of the bar immediately after string break?
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Thin uniform bar of mass m and length 2L is held at an angle son with ...
Analysis:
The given problem involves the application of Newton's laws of motion and torque to determine the tension in the string at the left end and the angular acceleration of the bar after the right string breaks.
Free Body Diagram:
- Draw a free body diagram of the bar showing all the forces acting on it.
- The forces acting on the bar are tension T at the left end, tension T' at the right end, weight mg acting at the center of the bar, and the reaction force R at the pivot point.
Equations of Motion:
- Apply Newton's second law in the vertical direction to determine the tension T.
- Apply the torque equation about the pivot point to determine the angular acceleration of the bar.
Equation for Tension:
- Resolving forces vertically, we get T - mgcosθ = 0, where θ is the angle of inclination.
- Solving for T, we get T = mgcosθ.
Equation for Angular Acceleration:
- Taking the torque about the pivot point, we get T' * 2L = I * α, where I is the moment of inertia of the bar about the pivot point and α is the angular acceleration.
- Substituting the values and solving, we get α = 3g / (2L).
Therefore, the tension in the string at the left end is T = mgcosθ and the angular acceleration of the bar immediately after the string break is α = 3g / (2L).
The given problem involves the application of Newton's laws of motion and torque to determine the tension in the string at the left end and the angular acceleration of the bar after the right string breaks.
Free Body Diagram:
- Draw a free body diagram of the bar showing all the forces acting on it.
- The forces acting on the bar are tension T at the left end, tension T' at the right end, weight mg acting at the center of the bar, and the reaction force R at the pivot point.
Equations of Motion:
- Apply Newton's second law in the vertical direction to determine the tension T.
- Apply the torque equation about the pivot point to determine the angular acceleration of the bar.
Equation for Tension:
- Resolving forces vertically, we get T - mgcosθ = 0, where θ is the angle of inclination.
- Solving for T, we get T = mgcosθ.
Equation for Angular Acceleration:
- Taking the torque about the pivot point, we get T' * 2L = I * α, where I is the moment of inertia of the bar about the pivot point and α is the angular acceleration.
- Substituting the values and solving, we get α = 3g / (2L).
Therefore, the tension in the string at the left end is T = mgcosθ and the angular acceleration of the bar immediately after the string break is α = 3g / (2L).
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Thin uniform bar of mass m and length 2L is held at an angle son with the horizontal by means of two vertical inextensible strings, at each end as shown in figure. If the string at the right end breaks, leaving the bar to swing, determine the tension in the string at the left end and the angular acceleration of the bar immediately after string break?
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Thin uniform bar of mass m and length 2L is held at an angle son with the horizontal by means of two vertical inextensible strings, at each end as shown in figure. If the string at the right end breaks, leaving the bar to swing, determine the tension in the string at the left end and the angular acceleration of the bar immediately after string break? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Thin uniform bar of mass m and length 2L is held at an angle son with the horizontal by means of two vertical inextensible strings, at each end as shown in figure. If the string at the right end breaks, leaving the bar to swing, determine the tension in the string at the left end and the angular acceleration of the bar immediately after string break? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Thin uniform bar of mass m and length 2L is held at an angle son with the horizontal by means of two vertical inextensible strings, at each end as shown in figure. If the string at the right end breaks, leaving the bar to swing, determine the tension in the string at the left end and the angular acceleration of the bar immediately after string break?.
Thin uniform bar of mass m and length 2L is held at an angle son with the horizontal by means of two vertical inextensible strings, at each end as shown in figure. If the string at the right end breaks, leaving the bar to swing, determine the tension in the string at the left end and the angular acceleration of the bar immediately after string break? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Thin uniform bar of mass m and length 2L is held at an angle son with the horizontal by means of two vertical inextensible strings, at each end as shown in figure. If the string at the right end breaks, leaving the bar to swing, determine the tension in the string at the left end and the angular acceleration of the bar immediately after string break? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Thin uniform bar of mass m and length 2L is held at an angle son with the horizontal by means of two vertical inextensible strings, at each end as shown in figure. If the string at the right end breaks, leaving the bar to swing, determine the tension in the string at the left end and the angular acceleration of the bar immediately after string break?.
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