Mimas, a moon of Saturn, has an orbital radius of 1.62 × 108 m and an orbital period of about 23.21 h. Use Newton’s version of K
1 answer:
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Answer:
Sphere, cylinder hoop
Explanation:
To analyze Which student is right it is best to propose the solution of the problem. Let's look for the speed of the center of mass. Let's use the concept of mechanical energy
In the highest part of the ramp
Em₀ = U = mg y
In the lowest part
Here the energy has part of translation and part of rotation
=
+
= ½ m
² + ½ I w²
Where I is the moment of inertia of the body and w the angular velocity that relates to the velocity of the center of mass
= w r
w = / r
Let's replace
= ½ I (
/ r)²
Energy is conserved
mg y = ½ m ² + ½ I
² / r2
½ (m + I / r²) ² = m g y
½ (1 + I / m r²) ² = g y
= √ [2gy / (1 + I / mr²)]
This is the velocity of the center of mass of the bodies, as they all have the same radius with comparing this point is sufficient. Now let's use the speed definition
v = d / t
t = d / v
t = d / (√ [2gy / (1 + I / mr²)])
t = (d / √ 2gy) √(1 + I / m r²)
Therefore we see that time is proportional to the square root. All quantities are constant and the one that varies is the moment of inertia.
The moments of inertia of
Sphere is Is = 2/5 M r²
Cylinder Ic = ½ M r²
Hoop Ih = M r²
Let's replace each one and calculate the time
Sphere
ts = (d / √2gy) √ (1 + 2/5 Mr² / mr²)
ts = (d / √ 2gy) √ (1 +2/5) = (d / √ 2gy) √(1.4)
ts = (d / √ 2gy) 1.1
Cylinder
tc = (d / √2gy) √ (1 + 1/2 Mr² / Mr²)
tc = (d / √2gy) √ (1 + ½) = (d / √ 2gy) √ 1.5
tc = (d / √ 2gy) 1.2
Hoop
th = (d / √2gy) √ (1 + mr² / mr²)
th = (d / √2gy) √(1 + 1) = (d / √ 2gy) √ 2
th = (d / √ 2gy) 1.41
We have the results for the time the body that arrives the fastest is the sphere and the one that is the most hoop. Therefore the correct answer is
ts < tc < th
Sphere, cylinder hoop