A uniform rod with a mass of m = 1.94 kg and a length of l = 2.10 m is attached to a horizontal surface with a hi=nge. The rod can rotate without friction. (See figure.)
Initially the rod is held at rest at an angle of Θ = 70.4 with respect to the horizontal surface. Then the rod is released.
What is the angular speed of the rod, when it lands on the horizontal surface?

What is the angular acceleration of the rod, just before it touches the horizontal surface?

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onyebuchinnaji onyebuchinnaji · Answer 1 · 2022-07-15T21:26:39+02:00

(a) The angular speed of the rod, when it lands on the horizontal surface is 1.61 rad/s.

(b) The the angular acceleration of the rod, just before it touches the horizontal surface is 1.06 rad/s² .

The angular acceleration of the rod is calculated as follows;

Apply the principle of conservation of angular momentum;

τ = Iα

where;

τ = Fr = mg(L/2) cosθ

I = mL²/3

mg(L/2) cosθ = mL²/3(α)

g(1/2) cosθ = L/3(α)

(³/₂)g cosθ = L(α)

(³/₂ L)g cosθ = α

1.5Lg cosθ = α

1.5(2.1) cos (70.4) = α

1.06 rad/s² = α

ωf² = ω₁² + 2αθ

ωf² = 0 + 2(1.06)(70.4π/180)

ωf² = 2.605

ωf = √2.605

ωf = 1.61 rad/s

Thus, the angular speed of the rod, when it lands on the horizontal surface is 1.61 rad/s.

The the angular acceleration of the rod, just before it touches the horizontal surface is 1.06 rad/s² .

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