<h2>K.E/P.E = m/k tan²φ x ω²</h2>
Explanation:
The given position of block x = x₀ cos(ωt + φ)
The velocity of block v = dx/dt = - x₀ sin(ωt + φ) x ω
The kinetic energy = 1/2 mv² = 1/2 m x₀² sin²(ωt + φ) x ω²
The potential energy of spring = 1/2 k x² , where k is the spring constant
Thus P.E = 1/2 x k x x₀² cos²(ωt + φ)
When t = 0
K.E = 1/2 m x₀²sin²φ x ω²
P.E = 1/2 k x₀² cos²φ
Dividing these , we have
K.E/P.E = m/k tan²φ x ω²
Answer:
a. 0.342 kg-m² b. 2.0728 kg-m²
Explanation:
a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m
I = 1/2MR²
= 1/2 × 56.5 kg × (0.11 m)²
= 0.342 kgm²
So the moment of inertia of the skater is
b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)
I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m
I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²
= 0.1802 kg-m² + 0.6852 kg-m²
= 0.8654 kg-m²
The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².
So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²
Shortwave radios have a higher frequency than ultraviolet light
The density is 21.6 / 2cm^3, so the density is 2.7 grams/cm^3