Question

A uniform rod is hinged as shown and is released from a horizontal position. The angular velocity of the rod as it passes the vertical position (neglect friction)

• A. $\sqrt{\frac{24g}{7l}}$
• B. $\sqrt{\frac{48 g}{8l}}$
• C. $\sqrt{\frac{12 g}{17ℓ}}$
• D. $\sqrt{\frac{3 g}{2l}}$

Answer 1

Answer: (A)

$\sqrt{\frac{24g}{7l}}$

Answer 2

To determine the angular velocity of the rod as it passes the vertical position, we use the principle of conservation of mechanical energy.

1. Define Initial and Final States:

• Initial State (Horizontal Position, released from rest):

  • Let the hinge be the reference level for potential energy (PE = 0).
  • The center of mass (CM) of the uniform rod is at its geometric center, l/2 from either end. The hinge is at l/4 from one end. Therefore, the distance of the CM from the hinge is d = l/2 - l/4 = l/4.
  • In the horizontal position, the CM is at the same height as the hinge.
  • Initial Potential Energy ($PE_i$) = 0.
  • Initial Kinetic Energy ($KE_i$) = 0 (since it's released from rest).
  • Total Initial Energy ($E_i$) = $PE_i + KE_i = 0$.

• Final State (Vertical Position):

  • When the rod is vertical, the CM is now l/4 below the hinge.
  • Final Potential Energy ($PE_f$) = -mg(l/4).
  • Final Kinetic Energy ($KE_f$) = (1/2)I_hinge ω², where I_hinge is the moment of inertia about the hinge and ω is the angular velocity.
  • Total Final Energy ($E_f$) = $PE_f + KE_f = -mg(l/4) + (1/2)I_hinge ω²$.

2. Calculate the Moment of Inertia about the Hinge ($I_{hinge}$):

• The moment of inertia of a uniform rod of mass m and length l about its center of mass ($I_{CM}$) is (1/12)ml².

• Using the parallel axis theorem, the moment of inertia about the hinge (which is at a distance d = l/4 from the CM) is:

  $I_{hinge} = I_{CM} + md²$

  $I_{hinge} = (1/12)ml² + m(l/4)²$

  $I_{hinge} = (1/12)ml² + m(l²/16)$

  To add these, find a common denominator (48):

  $I_{hinge} = (4/48)ml² + (3/48)ml²$

  $I_{hinge} = (7/48)ml²$

3. Apply Conservation of Mechanical Energy:

• Since friction is neglected, mechanical energy is conserved: $E_i = E_f$.

• $0 = -mg(l/4) + (1/2)I_{hinge} ω²$

• Rearrange the equation:

  $mg(l/4) = (1/2)I_{hinge} ω²$

4. Substitute $I_{hinge}$ and Solve for ω:

• $mg(l/4) = (1/2) * (7/48)ml² * ω²$

• Cancel mass m from both sides:

  $g(l/4) = (1/2) * (7/48)l² * ω²$

• Cancel one l from both sides (assuming l ≠ 0):

  $g/4 = (1/2) * (7/48)l * ω²$

• Multiply both sides by 2:

  $g/2 = (7/48)l * ω²$

• Isolate $ω²$:

  $ω² = (g/2) * (48 / (7l))$

  $ω² = (g * 48) / (2 * 7l)$

  $ω² = (g * 24) / (7l)$

  $ω² = \frac{24g}{7l}$

• Take the square root to find ω:

  $ω = \sqrt{\frac{24g}{7l}}$

The final answer is $\boxed{\sqrt{\frac{24g}{7l}}}$.

Explanation of the solution:

1. Energy Conservation: The rod is released from rest in a horizontal position. As it swings to the vertical position, its potential energy is converted into rotational kinetic energy.
2. Potential Energy Change: The center of mass (CM) of the rod moves from the hinge level (initial PE = 0) to a distance l/4 below the hinge (final PE = -mg(l/4)). The loss in potential energy is mg(l/4).
3. Rotational Kinetic Energy: The gain in kinetic energy is (1/2)Iω², where I is the moment of inertia about the hinge and ω is the angular velocity.
4. Moment of Inertia: For a uniform rod, $I_{CM} = (1/12)ml²$. The hinge is at l/4 from the CM. By the parallel axis theorem, $I_{hinge} = I_{CM} + m(l/4)² = (1/12)ml² + (1/16)ml² = (7/48)ml²$.
5. Equating Energies: mg(l/4) = (1/2) * (7/48)ml² * ω².
6. Solve for ω: Simplifying the equation yields ω = \sqrt{\frac{24g}{7l}}.

Answer:

The correct option is $\sqrt{\frac{24g}{7l}}$.