A rigid rod of length L has a mass per unit length that varies linearly with distance x from its left end, given by λ(x) = kx, where k is a positive constant. The rod is pivoted at its geometric center (L/2). To keep the rod in horizontal equilibrium, a mass M is hung from its left end (x=0). What mass M is required in terms of L and k?
Correct: C
To solve this problem, we first need to determine the total mass of the rod and its center of mass.
1. **Total Mass of the Rod (M_rod):**
M_rod = ∫₀ᴸ λ(x) dx = ∫₀ᴸ kx dx
M_rod = k [x²/2] from 0 to L
M_rod = kL²/2
2. **Center of Mass of the Rod (x_CM):**
x_CM = (∫₀ᴸ x λ(x) dx) / M_rod
∫₀ᴸ x λ(x) dx = ∫₀ᴸ x (kx) dx = k ∫₀ᴸ x² dx
k [x³/3] from 0 to L = kL³/3
x_CM = (kL³/3) / (kL²/2) = (kL³/3) * (2 / (kL²))
x_CM = 2L/3
So, the center of mass of the rod is at a distance of 2L/3 from its left end.
3. **Equilibrium Condition:**
The rod is pivoted at its center, which is at L/2 from the left end.
A mass M is hung from the left end (x=0).
Torques must balance about the pivot point (L/2).
- **Torque due to mass M (τ_M):**
The mass M is at x=0, which is L/2 distance to the left of the pivot.
τ_M = M * g * (L/2) (This creates a clockwise torque).
- **Torque due to the rod's weight (τ_rod):**
The rod's weight (M_rod * g) acts at its center of mass, x_CM = 2L/3.
The distance of x_CM from the pivot (L/2) is: d_rod = |x_CM - L/2| = |2L/3 - L/2| = |4L/6 - 3L/6| = L/6.
Since x_CM (2L/3) is to the right of the pivot (L/2), the rod's weight creates a counter-clockwise torque.
τ_rod = M_rod * g * (L/6)
For equilibrium, τ_M = τ_rod:
M * g * (L/2) = M_rod * g * (L/6)
M * (L/2) = M_rod * (L/6)
Substitute M_rod = kL²/2:
M * (L/2) = (kL²/2) * (L/6)
M * (L/2) = kL³/12
Solve for M:
M = (kL³/12) * (2/L)
M = kL²/6
The correct choice is C.