FTcose 1. For each of the four trials, determine the airplane's mass from Egn. (1)

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**Instructions for Airplane Motion Experiment**

1. **For each of the four trials, determine the airplane's mass** using the formula:
   \[
   m = \frac{F_T \cos \theta}{g}
   \]
   **Equation (1)**

2. **For each of the four trials:**
   - (a) **Find the period of the airplane’s motion**, which we will call \( T \).

   - (b) **Calculate the airplane’s mass** using the formula:
     \[
     m = \frac{(F_T \sin \theta) T^2}{4 \pi^2 r}
     \]
     **Equation (2)**

   In Equation (2), \( r \) is the radius of the circular path.

3. **Mass measurements of the airplane** (as measured on a scale):
   - Trial 1: 0.2935 kg
   - Trial 2: 0.2147 kg
   - Trial 3: 0.3935 kg
   - Trial 4: 0.2939 kg

4. **Calculating Percent Differences:**
   - (a) **Find the percent difference** between each of the masses given in step 3 and the corresponding mass calculated using Equation (1).
   - (b) **Find the percent difference** between each of the masses given in step 3 and the corresponding mass calculated using Equation (2).
Transcribed Image Text:**Instructions for Airplane Motion Experiment** 1. **For each of the four trials, determine the airplane's mass** using the formula: \[ m = \frac{F_T \cos \theta}{g} \] **Equation (1)** 2. **For each of the four trials:** - (a) **Find the period of the airplane’s motion**, which we will call \( T \). - (b) **Calculate the airplane’s mass** using the formula: \[ m = \frac{(F_T \sin \theta) T^2}{4 \pi^2 r} \] **Equation (2)** In Equation (2), \( r \) is the radius of the circular path. 3. **Mass measurements of the airplane** (as measured on a scale): - Trial 1: 0.2935 kg - Trial 2: 0.2147 kg - Trial 3: 0.3935 kg - Trial 4: 0.2939 kg 4. **Calculating Percent Differences:** - (a) **Find the percent difference** between each of the masses given in step 3 and the corresponding mass calculated using Equation (1). - (b) **Find the percent difference** between each of the masses given in step 3 and the corresponding mass calculated using Equation (2).
**Table: Experimental Data on Circular Motion**

| Trial | Diameter of Airplane’s Path | Tension in String | Angle of String | Time for Two Revolutions |
|-------|----------------------------|-------------------|----------------|--------------------------|
| 1     | 90.5 cm                    | 4.5 N             | 52             | 2.5000 sec               |
| 2     | 104.5 cm                   | 7.5 N             | 74             | 1.7333 sec               |
| 3     | 87.4 cm                    | 6.5 N             | 52             | 2.3917 sec               |
| 4     | 83.5 cm                    | 5.2 N             | 50             | 2.3250 sec               |

**Calculations:**

- Diameter for trial 1 is: \(75.5 \, \sin(52) = (45.31) \times 2 = 90.5 \, \text{cm}\)
- Diameter for trial 2 is: \(54.5 \, \sin(74) = (52.39) \times 2 = 104.5 \, \text{cm}\)
- Diameter for trial 3 is: \(55.5 \, \sin(52) = (43.73) \times 2 = 87.4 \, \text{cm}\)
- Diameter for trial 4 is: \(54.5 \, \sin(50) = (41.75) \times 2 = 83.5 \, \text{cm}\)

**Analysis:**

The table displays data collected from four trials investigating the relationship between the diameter of an airplane's path and various other factors in circular motion. Each trial records:

- **Diameter of Airplane’s Path:** The distance across the path, in centimeters (cm).
- **Tension in String:** Measured in Newtons (N).
- **Angle of String:** Measured in degrees.
- **Time for Two Revolutions:** The duration in seconds (sec) that it takes for two complete rotations.

The additional calculations show how the diameters for each trial are derived using trigonometric functions based on the sine of the given angles. This data is essential for analyzing the dynamics of rotational motion and the forces involved.
Transcribed Image Text:**Table: Experimental Data on Circular Motion** | Trial | Diameter of Airplane’s Path | Tension in String | Angle of String | Time for Two Revolutions | |-------|----------------------------|-------------------|----------------|--------------------------| | 1 | 90.5 cm | 4.5 N | 52 | 2.5000 sec | | 2 | 104.5 cm | 7.5 N | 74 | 1.7333 sec | | 3 | 87.4 cm | 6.5 N | 52 | 2.3917 sec | | 4 | 83.5 cm | 5.2 N | 50 | 2.3250 sec | **Calculations:** - Diameter for trial 1 is: \(75.5 \, \sin(52) = (45.31) \times 2 = 90.5 \, \text{cm}\) - Diameter for trial 2 is: \(54.5 \, \sin(74) = (52.39) \times 2 = 104.5 \, \text{cm}\) - Diameter for trial 3 is: \(55.5 \, \sin(52) = (43.73) \times 2 = 87.4 \, \text{cm}\) - Diameter for trial 4 is: \(54.5 \, \sin(50) = (41.75) \times 2 = 83.5 \, \text{cm}\) **Analysis:** The table displays data collected from four trials investigating the relationship between the diameter of an airplane's path and various other factors in circular motion. Each trial records: - **Diameter of Airplane’s Path:** The distance across the path, in centimeters (cm). - **Tension in String:** Measured in Newtons (N). - **Angle of String:** Measured in degrees. - **Time for Two Revolutions:** The duration in seconds (sec) that it takes for two complete rotations. The additional calculations show how the diameters for each trial are derived using trigonometric functions based on the sine of the given angles. This data is essential for analyzing the dynamics of rotational motion and the forces involved.
Expert Solution
Given

Diameter of airplane's path 1, D1=90.5cm=90.5×10-2mDiameter of airplane's path 2, D2=104.5cm=104.5×10-2mDiameter of airplane's path 3, D3=87.4cm=87.4×10-2mDiameter of airplane's path 4, D4=83.5cm=83.5×10-2mTension of 1, FT1=4.5NTension of 2, FT2=7.5NTension of 3, FT3=6.5NTension of 4, FT4=5.2NAngle of 1, θ1=520Angle of 2, θ2=740Angle of 3, θ3=520Angle of 4, θ4=500Time for two revolutions of 1, T1=2.5000sTime for two revolutions of 2, T2=1.7333sTime for two revolutions of 3, T3=2.3917sTime for two revolutions of 4, T4=2.3250sAcceleration due to gravity, g=9.8m/s2

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