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umka2103 [35]
3 years ago
11

Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m

ass of 5.00×10-2 {kg}.1. About an axis perpendicular to the rod and passing through its center. (in I=kg*m^2)2. About an axis perpendicular to the rod and passing through one end.3. About an axis along the length of the rod.
Physics
1 answer:
mamaluj [8]3 years ago
3 0

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

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- The mass of passenger, mp = 70 kg

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- The initial speed of the bug, vbi = 1.5 m/s

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a.What is the initial momentum of you plus your bicycle?

b.What is the initial momentum of the bug?

c.What is your change in velocity due to the collision the bug?

d.What would the change in velocity have been if the bug were traveling in the opposite direction?

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- First we will set our one dimensional coordinate system, taking right to be positive in the direction of bicycle.

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                       Pi,c = vpi* ( mc + mp)

                       Pi,c = 13* ( 12+ 70 )

                       Pi,c = 1066 kgm/s  

- The initial linear momentum (Pi,b) of the bug would be:

                       Pi,b = vbi*mb

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                       Pi,b = 0.0075 kgm/s  

- We will consider the bicycle, the passenger and the bug as a system in isolation on which no external unbalanced forces are acting. This validates the use of linear conservation of momentum.

- The bicycle, passenger and bug all travel in the (+x) direction after the bug splatters on the helmet.

                       Pi = Pf

                       Pi,c + Pi,b = V*(mb + mc + mp)

Where,    V : The velocity of the (bicycle, passenger and bug) after collision.

                      1066 + 0.0075 = V*( 0.005 + 12 + 70 )

                      V = 1066.0075 / 82.005

                      V = 12.9993 m/s

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- We will apply the linear conservation of momentum similarly.

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                      Pi,c + Pi,b = V*(mb + mc + mp)        

                      1066 - 0.0075 = V*( 0.005 + 12 + 70 )

                      V = 1065.9925 / 82.005

                      V = 12.99911 m/s

- The change in velocity is Δv = 13 - 12.99911 =  -0.00088 m/s      

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