a) What is the sphere’s angular velocity at the bottom of the incline?
b) What fraction of its kinetic energy is rotational?
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, therefore
mg (h sinθ) = 1/2 (2/5 mr²) ω² + 1/2 mv², now we can eliminate the mass
g h sin θ = 1/5 r² ω² + 1/2 v²
ω = v/r, v = ωr
consequently,
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