A spherical container has a radius of 0.2 m and mass of 10 kg. the two spouts ca

Question

A spherical container has a radius of 0.2 m and mass of 10 kg. the two spouts can be considered massless but extend an additional 0.10 m above the surface of the container. The container is hollow. Do not consier the moment of inertia of the steam inside the container

1) What is the moment of inertia of this container?

2) if the pressure inside the container is 1.5 atmosphere, what is the angular acceleration of the container? take the area of spouts to be circle of radius 0.010m. The outside pressure is 1 atomsphere

3)how much energy is stored by the steam inside the container

4) if the energy is somehow converted into into rotational kinetic energy, what would be the final angular velocity of the container?

Explanation / Answer

a. I = (2/3)MR^2 because the problem assumes that the two spouts are massless. This gives me 0.27 kg * m^2

b. P = F/A
F = (1.5 atm - 1.0 atm)(pi(0.010m)^2)
Convert 0.5 atm to Pascals --> 50,662.5 Pa.
F = 15.9 N

Net Torque = 2(15.9N)(0.30m) = 9.5 N * m = Ia
Angular Acceleration = 9.5 N *m / 0.27 kg *m^2 = 35 rad/s^2

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