Answer:
(b) 0.0176
(c) -0.0124
(d) 209
(e) Also comes to rest
(a) 2.38
(b) 5.95
Explanation:
(a) Your answer is correct. Angular momentum is conserved, so as the lighter beetle moves clockwise, the heavier turntable will move counterclockwise at a slower speed.
(b/c) Initial angular momentum = final angular momentum
L₀ = L
I₁ ω₁,₀ + I₂ ω₂,₀ = I₁ ω₁ + I₂ ω₂
0 = mr² ω₁ + ½ Mr² ω₂
0 = 2m ω₁ + M ω₂
The beetle's angular velocity relative to the turntable is 0.03 rad/s, so ω₁ = ω₂ + 0.03. Plugging in:
0 = 2 (30 g) (ω₂ + 0.03 rad/s) + (85 g) ω₂
0 = 60ω₂ + 1.8 rad/s + 85ω₂
145ω₂ = -1.8 rad/s
ω₂ = -0.0124 rad/s
ω₁ = ω₂ + 0.03
ω₁ = 0.0176 rad/s
Relative to a stationary observer, the beetle moves 0.0176 rad/s clockwise and the turntable moves -0.0124 rad/s counterclockwise.
(d) Angular distance = angular velocity × time
2π rad = (0.03 rad/s) t
t = 209 s
(e) Angular momentum is conserved. Since both the beetle and the turntable were originally at rest, the turntable will again come to rest when the beetle stops.
(a) Angular momentum is conserved.
L₀ = L
I₀ ω₀ + I₂ ω₀ = I ω + I₂ ω
(I₀ + I₂) ω₀ = (I + I₂) ω
(M (R/2)² + ½ (3M) (R)²) ω₀ = (M (R)² + ½ (3M) (R)²) ω
(¼ MR² + ³/₂ MR²) ω₀ = (MR² + ³/₂ MR²) ω
(¼ + ³/₂) ω₀ = (1 + ³/₂) ω
(1 + 6) ω₀ = (4 + 6) ω
7ω₀ = 10ω
ω = 0.7ω₀
ω = 0.7 (3.40 rad/s)
ω = 2.38 rad/s
(b) Angular momentum is conserved.
L₀ = L
I₀ ω₀ + I₂ ω₀ = I ω
(I₀ + I₂) ω₀ = I ω
(M (R/2)² + ½ (3M) (R)²) (3.40 rad/s) = M (R)² ω
(¼ MR² + ³/₂ MR²) (3.40 rad/s) = MR² ω
(¼ + ³/₂) (3.40 rad/s) = ω
ω = 5.95 rad/s
Notice we could also have used our answer from part a and I₀ = MR².
(I₀ + I₂) ω₀ = I ω
(M (R)² + ½ (3M) (R)²) (2.38 rad/s) = M (R)² ω
(MR² + ³/₂ MR²) (2.38 rad/s) = MR² ω
(1 + ³/₂) (2.38 rad/s) = ω
ω = 5.95 rad/s
