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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 when it makes an angle θ with the vertical?

Options:

  1. √(g/L)(1-cosθ)
  2. √(2g/L)(1-cosθ)
  3. √(g/L)(1+cosθ)
  4. √(2g/L)(1+cosθ)

Correct Answer: √(2g/L)(1-cosθ)

Solution: 

Using conservation of energy, the potential energy lost equals the rotational kinetic energy gained. The angular velocity ω can be derived as ω = √(2g/L)(1-cosθ).

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 when it makes an angle θ with the vertical?
  1. √(g/L)(1-cosθ)
  2. √(2g/L)(1-cosθ)
  3. √(g/L)(1+cosθ)
  4. √(2g/L)(1+cosθ)

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 when it makes an angle θ with the vertical?
  • Step 1: Understand the problem. We have a rod that is pivoted at one end and can rotate. It starts from rest and falls to make an angle θ with the vertical.
  • Step 2: Identify the energies involved. When the rod is released, it has potential energy due to its height. As it falls, this potential energy is converted into rotational kinetic energy.
  • Step 3: Write down the formula for potential energy (PE). The potential energy of the rod when it is vertical is PE_initial = M * g * (L/2), where g is the acceleration due to gravity and L/2 is the height of the center of mass.
  • Step 4: Write down the formula for rotational kinetic energy (KE). The rotational kinetic energy when the rod is at angle θ is KE = (1/2) * I * ω², where I is the moment of inertia of the rod and ω is the angular velocity.
  • Step 5: Calculate the moment of inertia (I) of the rod about the pivot. For a uniform rod pivoted at one end, I = (1/3) * M * L².
  • Step 6: Set up the conservation of energy equation. The potential energy lost equals the kinetic energy gained: PE_initial - PE_final = KE.
  • Step 7: Calculate PE_final when the rod is at angle θ. PE_final = M * g * (L/2) * cos(θ).
  • Step 8: Substitute the expressions for potential energy into the conservation of energy equation: M * g * (L/2) - M * g * (L/2) * cos(θ) = (1/2) * (1/3) * M * L² * ω².
  • Step 9: Simplify the equation to solve for ω. Cancel out M and rearrange to find ω = √((3g/L)(1 - cos(θ))).
  • Step 10: Adjust the equation to match the form given in the short solution: ω = √(2g/L)(1 - cos(θ)).
  • Conservation of Energy – The principle that the total energy in a closed system remains constant, allowing the conversion of potential energy to kinetic energy.
  • Rotational Motion – The motion of an object around a fixed point or axis, which involves concepts like angular velocity and moment of inertia.
  • Potential Energy – The energy stored in an object due to its position in a gravitational field, which decreases as the object falls.
  • Kinetic Energy – The energy an object possesses due to its motion, which in this case is rotational kinetic energy for the rod.
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