A particle performing uniform circular motion has angular momentum L. What will be the new angular momentum, if its angular freq
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Answer:
The final translational seed at the bottom of the ramp is approximately 4.84 m/s
Explanation:
The given parameters are;
The radius of the sphere, R = 2.50 cm
The mass of the sphere, m = 4.75 kg
The translational speed at the top of the inclined plane, v = 3.00 m/s
The length of the inclined plane, l = 2.75 m
The angle at which the plane is tilted, θ = 22.0°
We have;
+
=
+
K = (1/2)×m×v²×(1 + I/(m·r²))
I = (2/5)·m·r²
K = (1/2)×m×v²×(1 + 2/5) = 7/10 × m×v²
U = m·g·h
h = l×sin(θ)
h = 2.75×sin(22.0°)
∴ 7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) = 7/10 × 4.75ײ + 0
7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) ≈ 77.93
∴ 77.93 ≈ 7/10 × 4.75ײ
² = 77.93/(7/10 × 4.75)
≈ √(77.93/(7/10 × 4.75)) ≈ 4.84
The final translational seed at the bottom of the ramp, ≈ 4.84 m/s.