Math Problem Statement

Calculate the rotational inertia (I) of a solid disk and a ring, using the provided measurements and formulas.

Solution

The provided image outlines a calculation of rotational inertia for both a solid disk and a ring. Here’s a step-by-step explanation of the calculations shown:

1. Rotational Inertia of the Disk (IdiskI_\text{disk})

The formula for the rotational inertia of a solid disk is: Idisk=12MR2I_\text{disk} = \frac{1}{2} M R^2 Where:

  • M=1425g=1.425kgM = 1425 \, \text{g} = 1.425 \, \text{kg} (mass of the disk)
  • R=diameter of the disk2=22.82cm2=11.41cm=0.1141mR = \frac{\text{diameter of the disk}}{2} = \frac{22.82 \, \text{cm}}{2} = 11.41 \, \text{cm} = 0.1141 \, \text{m}

Calculation: Idisk=12×1.425×(0.1141)2I_\text{disk} = \frac{1}{2} \times 1.425 \times (0.1141)^2

Converting back to gcm2\text{g} \cdot \text{cm}^2 as seen in the image: Idisk=12×1425×(11.41)2I_\text{disk} = \frac{1}{2} \times 1425 \times (11.41)^2 Idisk=92,759gcm2I_\text{disk} = 92,759 \, \text{g} \cdot \text{cm}^2

2. Rotational Inertia of the Ring (IringI_\text{ring})

The formula for the rotational inertia of a ring is: Iring=12M(Rout2+Rin2)I_\text{ring} = \frac{1}{2} M \left( R_\text{out}^2 + R_\text{in}^2 \right) Where:

  • M=1442g=1.442kgM = 1442 \, \text{g} = 1.442 \, \text{kg} (mass of the ring)
  • Rout=outer diameter2=12.63cm2=6.315cmR_\text{out} = \frac{\text{outer diameter}}{2} = \frac{12.63 \, \text{cm}}{2} = 6.315 \, \text{cm}
  • Rin=inner diameter2=10.75cm2=5.375cmR_\text{in} = \frac{\text{inner diameter}}{2} = \frac{10.75 \, \text{cm}}{2} = 5.375 \, \text{cm}

Calculation: Iring=12×1442×((6.315)2+(5.375)2)I_\text{ring} = \frac{1}{2} \times 1442 \times \left( (6.315)^2 + (5.375)^2 \right) Breaking down the radii terms: Rout2=(6.315)2=39.87cm2,Rin2=(5.375)2=28.89cm2R_\text{out}^2 = (6.315)^2 = 39.87 \, \text{cm}^2, \quad R_\text{in}^2 = (5.375)^2 = 28.89 \, \text{cm}^2 Adding these together: Rout2+Rin2=39.87+28.89=68.76cm2R_\text{out}^2 + R_\text{in}^2 = 39.87 + 28.89 = 68.76 \, \text{cm}^2

Substituting back: Iring=12×1442×68.76I_\text{ring} = \frac{1}{2} \times 1442 \times 68.76 Iring=49,811gcm2I_\text{ring} = 49,811 \, \text{g} \cdot \text{cm}^2


Let me know if you'd like further clarification or help understanding the process!

Here are some related questions to explore:

  1. How do uncertainties in the measurements affect the calculated values of rotational inertia?
  2. What are the physical implications of the calculated values of IdiskI_\text{disk} and IringI_\text{ring}?
  3. Can these formulas for rotational inertia be extended to 3D objects of different shapes?
  4. How does rotational inertia relate to angular acceleration and torque?
  5. What real-world applications might involve using these specific calculations?

Tip: Always check units carefully when performing physics calculations, especially when converting between metric units!

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Math Problem Analysis

Mathematical Concepts

Rotational Inertia
Physics of Motion
Geometry

Formulas

I_disk = (1/2) * M * R^2
I_ring = (1/2) * M * (R_out^2 + R_in^2)

Theorems

Moment of Inertia Principles

Suitable Grade Level

Grades 11-12