A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.7 m and moment of inertia 890 kg⋅m
1 answer:
Answer:
Explanation:
Using the Condition That Initial Angular Velocity is equal to final angular velocity
total angular momentum is equal to angular momentum of the person + angular Momentum of the Platform
Note L= I ×ω
Final Angular momentum of the person is equal to the final angular momentum of the platform
Final Moment of Inertia of the person I_{per.f} = =70×(2.7)²=510.3Kgm2
Initial Angular Momentum L_{i} = Final Angular Momentum L_{f}
I_{plat.i} ×ω_{plat.i} + I_{per.i} ×ω_{per.i} = I_{plat.f} ×ω_{plat.f} + I_{per.f} ×ω_{per.f}
890(0.85) + 0 =890(ω_{plat.f}) +510.3(ω_{plat.f})
756.5 = 1400.3 (ω_{plat.f})
(ω_{plat.f}) =0.54rad/s
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Answer:
Capacitance of the second capacitor = 2C
Explanation:
Where A is the area, d is the gap between plates and ε₀ is the dielectric constant.
Let C₁ be the capacitance of first capacitor with area A₁ and gap between plates d₁.
We have
Similarly for capacitor 2
Capacitance of the second capacitor = 2C
Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as
We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.
The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is
where ‘h’ is the length of the imaginary Gaussian surface.
2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,
3. At the boundary where r = R:
As can be seen from above, two E-field values are equal as predicted.