In the following problems, assume the following parameters:

Mass of the bottom steel disc: 1.468 kg.
Mass of the upper steel disc: 1.479 kg.
Mass of the upper aluminum disc: 0.461 kg.
Radius of each disc: 63.3 mm.
Radius of the take-up spool: r= 1.25 cm.
Mass of the falling weight: 25 grams.1.

1. Using the values given above, calculate the following:

a. The moment of inertia of the lower, steel disc.
b. The moment of inertia of the upper, steel disc.
c. The moment of inertia of the upper, aluminum disc.
d. The combined moment of inertia of the two steel discs.

2. Using the results of problem 1, what percentage error is made by neglecting mr2with respect to Iin equation (95)?
3. Suppose that the counter is reading 400 counts per second at some moment. What is the value of Ïat that moment?

Question

Babydes7247go Babydes7247go

Physics

Answered

In the following problems, assume the following parameters:

Mass of the bottom steel disc: 1.468 kg.
Mass of the upper steel disc: 1.479 kg.
Mass of the upper aluminum disc: 0.461 kg.
Radius of each disc: 63.3 mm.
Radius of the take-up spool: r= 1.25 cm.
Mass of the falling weight: 25 grams.1.

1. Using the values given above, calculate the following:

a. The moment of inertia of the lower, steel disc.
b. The moment of inertia of the upper, steel disc.
c. The moment of inertia of the upper, aluminum disc.
d. The combined moment of inertia of the two steel discs.

2. Using the results of problem 1, what percentage error is made by neglecting mr2with respect to Iin equation (95)?
3. Suppose that the counter is reading 400 counts per second at some moment. What is the value of Ïat that moment?

Answer :

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Moya1316go Moya1316go · Answer 1

Answer:

1) a) I₁ = 0.2941 kg m², b) I₂ = 0.2963 kg m², c) I_{total} = 0.5904 kg m²

3) α = 6.31 10⁶ rad / s²

Explanation:

1) The moment of inertia for bodies with high symmetry is tabulated, for a divo with an axis passing through its center is

I = ½ m r²

a) moment of inertia of the upper disk

I₁ = ½ m₁ r₁²

I₁ = ½ 1,468 0.633²

I₁ = 0.2941 kg m²

b) upper aluminum disc moment of inertia

I₂ = ½ m₂ r₂²

I₂ = ½ 1.479 0.633²

I₂ = 0.2963 kg m²

c) the moment of inertia is an additive scalar quantity therefore

I_{total} = I₁ + I₂

I_{total} = 0.2941 + 0.2963

I_{total} = 0.5904 kg m²

3) ask the value of the angular acceleration, that is, the second derivative of the angle with respect to time squared

indicate the angular velocity of the system w = 400 rev / s

Let's reduce the SI system

w = 400 rev / s (2π rad / rev) = 2513.27 rad / s

as the system is rotating we can calculate the centripetal acceleration

a = w² R

a = 2513.27² 0.633

a = 3.998 10⁶ m / s²

the linear and angular variable are related

a = α r

α = a / r

α = 3.998 10⁶ / 0.633

α = 6.31 10⁶ rad / s²