rigid body & gas law

(1) Using the parallel axis theorem, we have:
Moment of inertia about the edge of a solid cylinder with mass m and radius r:
mr2/2 + mr2 = 3mr2/2
Moment of inertia about the edge of a hollow cylinder with mass m and radius r:
mr2 + mr2 = 2mr2
So in case of solid cylinder:
Moment of weight component mg sin θ about the contact point = mgr sin θ, where θ is the inclination of the plane.
So angular acceleration for:
Solid cylinder = mgr sin θ/(3mr2/2) = (2 g sin θ)/(3r)
Hollow cylinder = mgr sin θ/(2mr2) = (g sin θ)/(2r)
Then, linear acceleration for:
Solid cylinder = (2 g sin θ)/(3r) x r = (2 g sin θ)/3
Hollow cylinder = (g sin θ)/(2r) x r = (g sin θ)/2
Hence, for any solid cylinder, the linear acceleration will be the same and so, II and III will reach the bottom at the same time which is faster than the hollow one.
(2) Using the equation PV = nRT, with n = 1, R = 8.314, P = 105 and T = 273, we can find out V = 0.02270 m3 which is the molar volume of any ideal gas under s.t.p.
Then, molar mass of nitrogen using this data will be:
1.25 x 0.02270 = 0.02837 kg/mol
So under 77 C, i.e. 350 K, by the formula:
vrms = √(3RT/M) where M = molar mass
We have:
vrms = 555 m/s

2009-05-21 17:19:08 補充：
For Q1, with diagram:
http://i388.photobucket.com/albums/oo325/loyitak1990/May09/Crazymech2.jpg

In fact we just need to who have a faster acceleration since all of them start from rest.

1. From the equation of motion: s = (1/2)(u+v).t
i.e. t = 2s/(u+v), since u = 0 m/s in this example, t = 2s/v i.e. for a given distance s, a higher velocity v of the cylinder at the bottom requred less time t.
Let's then find the velocities of the three cylinders at the bottom. Using conservation of energy,
MgH = (1/2).M.v^2 + (1/2).I.w^2
where g is the acceleration due to gravity
I is the moment of inertia of the cylinder
w is the angular velocity of the cylinder
But w = v/r, where r is the radius of the cylinder
hence, MgH = (1/2).M.v^2 + (1/2).I.v^2/r^2 = (1/2)(v^2).[M + I/r^2]
For the hollow cylinder with radius R, I = MR^2, r = R
thus, MgH = (1/2)v^2[M + M] = Mv^2
v^2 = gH
For the solid cylinder of radius R/2, I = (1/2).M(R/2)^2, r = R/2
thus, MgH = (1/2)v^2[M + (1/2)M] = (3/4).Mv^2
i.e. v^2 = (4/3)gH
For the solid vylinder of radius R, I = (1/2).M.R^2, r = R
thus MgH = (1/2)v^2[M + (1/2)M] = (3/4)Mv^2
i.e. v^2 = (4/3)gH
It can be seen that both cylinder II and III have the same velocity at bottom, which is higher than that of the hollow cylinder. Thus the time needed to travel from release to the bottom for cylinders II and III is the least.
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2. Using the relation: p = (1/3).d.c^2
where p is the gas pressure, d is the density of gas, c is the rms speed of gas molecules
i.e. c^2 = 3p/d = 3 x 10^5/1.25 (m/s)^2
or rms speed at 273 K, c = square-root[3 x 10^5/1.25] m/s = 489.9 m/s
Since the rms speed is proportional to the square-root of absolute temperature of the gas [for proof, please refer to any standard physics textbooks], hence
c(at 77 C) = 489.9 x square-root[(77+273)/273] m/s = 555 m/s