Respuesta :
Answer:
a) 88.1 rad/s
b) 0.286
Explanation:
given information:
diameter, d = 8 cm = 0.08 m
sphere's mass, m = 400 g = 0.4 kg
the distance from rest to the tip, h = 2.1 m
incline angle, θ = 25°
a) What is the sphere's angular velocity at the bottom of the incline?
mg(h sinθ) = 1/2 Iω² + 1/2mv²
I of solid sphere = 2/5 mr², thus
mg(h sinθ) = 1/2 (2/5 mr²) ω² + 1/2 mv², now we can remove the mass
g h sin θ = 1/5 r² ω² + 1/2 v²
ω = v/r, v = ωr
so,
g h sin θ = 1/5 r² ω² + 1/2 (ωr)²
g h sin θ = (7/10) r² ω²
ω² = 10 g h sin θ/7 r²
ω = √10 g h sin θ/7 r²
= √10 (9.8) (2.1) sin 25° / 7 (0.04)²
= 88.1 rad/s
b) What fraction of its kinetic energy(KE) is rotational?
fraction of its kinetic energy = rotational KE / total KE
total KE = total potential energy
= m g h sin θ
= 0.4 x 9.8 x 2.1 sin 25°
= 3.48 J
rotational KE = 1/2 Iω²
= 1/5 mr²ω²
= 1/5 0.4 (0.04)²(88.1)²
= 0.99
fraction of its KE = 0.99/3.48
= 0.286
A) The sphere's angular velocity at the bottom of the incline is; ω = 88.1 rad/s
B) Fraction of its kinetic energy that is rotational is; 0.286
What is the angular velocity?
We are given;
Diameter; d = 8 cm = 0.08 m
Mass of sphere; m = 400 g = 0.4 kg
Distance from rest to the tip; h = 2.1 m
Angle of inclination; θ = 25°
a) To get the sphere's angular velocity at the bottom of the incline, we will use the expression;
mg(h*sinθ) = ¹/₂Iω² + ¹/₂mv²
where;
I of solid sphere = ²/₅mr²
Thus;
mg(h*sinθ) = ¹/₂(²/₅mr²)ω² + ¹/₂mv²
The mass m will cancel out to give;
gh*sin θ = ¹/₅r²ω² + ¹/₂v²
where v = ωr
Thus;
gh*sin θ = ¹/₅r²ω² + ¹/₂r²ω²
gh*sin θ = ⁷/₁₀r²ω²
ω = √[(¹⁰/₇)*g*h*(sin θ)/r²]
ω = √[(¹⁰/₇)*9.8*2.1*(sin 25)/(0.04)²]
ω = 88.1 rad/s
b) Fraction of its kinetic energy that is rotational = rotational KE/total KE
But, total KE = total potential energy
Thus;
KE_tot = mgh*sin θ
KE_tot = 0.4 * 9.8 * 2.1 sin 25°
KE_tot = 3.48 J
KE_rot = ¹/₂Iω²
I of solid sphere = ²/₅mr². Thus;
KE_rot = ¹/₅mr²ω²
KE_rot = ¹/₅ * 0.4 * 0.04² * 88.1²
KE_rot = 0.99 J
Fraction of its kinetic energy that is rotational = 0.99/3.48 = 0.286
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