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A uniform rod of length L and mass M is pivoted at one end and released from res

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Question: A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular velocity of the rod just before it hits the ground? (2019)

Options:

  1. √(3g/L)
  2. √(2g/L)
  3. √(g/L)
  4. √(4g/L)

Correct Answer: √(2g/L)

Exam Year: 2019

Solution: 

Using conservation of energy, potential energy at the top is converted to rotational kinetic energy at the bottom. The angular velocity Ο‰ can be found using the relation Ο‰ = √(3g/L).

A uniform rod of length L and mass M is pivoted at one end and released from res

Practice Questions

Q1
A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular velocity of the rod just before it hits the ground? (2019)
  1. √(3g/L)
  2. √(2g/L)
  3. √(g/L)
  4. √(4g/L)

Questions & Step-by-Step Solutions

A uniform rod of length L and mass M is pivoted at one end and released from rest. What is the angular velocity of the rod just before it hits the ground? (2019)
  • Step 1: Understand that the rod is pivoted at one end and will rotate downwards due to gravity.
  • Step 2: Recognize that when the rod is at the top, it has potential energy because of its height.
  • Step 3: When the rod falls and reaches the bottom, all that potential energy is converted into rotational kinetic energy.
  • Step 4: Use the formula for potential energy (PE = M * g * h) where h is the height of the center of mass of the rod when it is vertical. The height h is L/2.
  • Step 5: Calculate the potential energy at the top: PE = M * g * (L/2).
  • Step 6: The rotational kinetic energy (KE) at the bottom is given by the formula KE = (1/2) * I * ω², where I is the moment of inertia of the rod about the pivot point.
  • Step 7: For a uniform rod pivoted at one end, the moment of inertia I = (1/3) * M * LΒ².
  • Step 8: Set the potential energy equal to the rotational kinetic energy: M * g * (L/2) = (1/2) * (1/3) * M * LΒ² * ω².
  • Step 9: Simplify the equation to solve for ω²: g * (L/2) = (1/6) * LΒ² * ω².
  • Step 10: Rearrange the equation to find Ο‰: ω² = (3g) / L.
  • Step 11: Take the square root to find the angular velocity: Ο‰ = √(3g/L).
  • Conservation of Energy – The principle that the total energy in a closed system remains constant, allowing potential energy to convert into kinetic energy.
  • Rotational Kinetic Energy – The energy possessed by a rotating object, which can be expressed in terms of angular velocity.
  • Moment of Inertia – A measure of an object's resistance to changes in its rotation, which is relevant when calculating angular velocity.
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