When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____
1 answer:
Answer and explanation:
When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.
The total car tire diameter can be calculated as:
The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:
pr=0.95·75=71.25
The problem is that the profiles are normalized and the nearest profile available is 70.
If we take a theorical tire with a profile of 71.25:
The theorical tire size should be 205/71 R15.
If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.
The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').
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Answer:
A) = 1.44 kg m², B) moment of inertia must increase
Explanation:
The moment of inertia is defined by
I = ∫ r² dm
For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is
I = ½ m R²
A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is
I = + m D²
Let's apply these equations to our case
The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms
=
+ 2
= ½ M R²
The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body
M = 7/8 m total
M = 7/8 64
M = 56 kg
The mass of the arms is
m’= 1/8 m total
m’= 1/8 64
m’= 8 kg
As it has two arms the mass of each arm is half
m = ½ m ’
m = 4 kg
The arms are very thin, we will approximate them as a particle
= M D²
Let's write the equation
= ½ M R² + 2 (m D²)
Let's calculate
= ½ 56 0.20² + 2 4 0.20²
= 1.12 + 0.32
= 1.44 kg m²
b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase