Answer:
6.9 m/s
v = √(3gL (1 − cos θ))
Explanation:
As the stick falls, its gravitational potential energy (PE) is converted to rotational kinetic energy (RE). Initially, when the stick is standing vertically, it has only potential energy. When the stick lands, its height is zero, so it has no potential energy, only kinetic energy.
PE₀ = RE
Gravitational potential energy is equal to weight (mg) times height (h) of the center of mass. Rotational kinetic energy is half the moment of inertia (I) times the square of the angular speed (ω).
mgh₀ = ½ Iω²
The center of mass is at the middle of the stick, so h₀ = L/2. For a stick rotating about its end, the moment of inertia is a third of the mass (m) times the square of the length (L). Since the end of the stick on the ground doesn't slip, the angular speed of the stick is equal to the linear speed (v) of the other end divided by the stick's length (L).
mg (L/2) = ½ (⅓ mL²) (v/L)²
½ mgL = ⅙ mv²
v² = 3gL
Plug in values and solve.
v² = 3 (9.8 m/s²) (1.6 m)
v² = 47.04 m²/s²
v = 6.86 m/s
Rounding to two significant figures, the speed of the end of the stick is 6.9 m/s.
Next, when the stick has fallen to an angle θ, it will have some potential energy.
PE₀ = PE + RE
mgh₀ = mgh + ½ Iω²
Using trigonometry, we can show that h = L/2 cos θ. The other formulas are the same as before.
mg (L/2) = mg (L/2 cos θ) + ½ (⅓ mL²) (v/L)²
½ mgL = ½ mgL cos θ + ⅙ mv²
½ mgL (1 − cos θ) = ⅙ mv²
3gL (1 − cos θ) = v²
v = √(3gL (1 − cos θ))
