Activity: Uniform Circular Motion - Swinging the Pipes, demonstrating the motion of a bob on a string.

Let’s solve this step by step.

We are told:
- There are 4 bobs (T-fittings) with different string lengths → so 4 different radii: r₁, r₂, r₃, r₄.
- Mass of each bob = 0.070 kg (same for all).
- For each bob, we swing it 20 times and record the time t (in seconds).
- We need to calculate centripetal force F for each bob using the formulas given.

But — wait! The problem doesn’t give us actual measured times or radii values. That means this worksheet is meant to be filled out after doing the experiment. Since you’re asking me to “solve” it, I’ll assume you want a sample calculation using made-up but realistic data to show how it works.

Let’s pretend we did the experiment and got these results:

| Bob # | Radius r (m) | Time for 20 swings t (s) |
|-------|--------------|--------------------------|
| 1 | 0.30 | 18.0 |
| 2 | 0.60 | 25.0 |
| 3 | 0.90 | 31.0 |
| 4 | 1.20 | 36.0 |

*(Note: These numbers are invented for demonstration. In real life, you’d measure them.)*

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Step-by-step for Bob #1 (shortest):

Given:
- m = 0.070 kg
- r = 0.30 m
- n = 20 rotations
- t = 18.0 s

Step 1: Circumference C = 2·π·r

C = 2 × π × 0.30 ≈ 2 × 3.1416 × 0.30 ≈ 1.885 m

Step 2: Rotation speed v = (n · C) / t

v = (20 × 1.885) / 18.0 ≈ 37.7 / 18.0 ≈ 2.094 m/s

Step 3: Centripetal acceleration a_c = v² / r

a_c = (2.094)² / 0.30 ≈ 4.385 / 0.30 ≈ 14.617 m/s²

Step 4: Centripetal force F = m · a_c

F = 0.070 × 14.617 ≈ 1.023 N → rounded to third decimal place: 1.023 N

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Bob #2 (medium):

r = 0.60 m, t = 25.0 s

C = 2 × π × 0.60 ≈ 3.770 m
v = (20 × 3.770) / 25.0 = 75.4 / 25.0 = 3.016 m/s
a_c = (3.016)² / 0.60 ≈ 9.096 / 0.60 ≈ 15.160 m/s²
F = 0.070 × 15.160 ≈ 1.061 N → 1.061 N

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Bob #3 (long):

r = 0.90 m, t = 31.0 s

C = 2 × π × 0.90 ≈ 5.655 m
v = (20 × 5.655) / 31.0 ≈ 113.1 / 31.0 ≈ 3.648 m/s
a_c = (3.648)² / 0.90 ≈ 13.308 / 0.90 ≈ 14.787 m/s²
F = 0.070 × 14.787 ≈ 1.035 N → 1.035 N

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Bob #4 (longest):

r = 1.20 m, t = 36.0 s

C = 2 × π × 1.20 ≈ 7.540 m
v = (20 × 7.540) / 36.0 ≈ 150.8 / 36.0 ≈ 4.189 m/s
a_c = (4.189)² / 1.20 ≈ 17.548 / 1.20 ≈ 14.623 m/s²
F = 0.070 × 14.623 ≈ 1.024 N → 1.024 N

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Wait — let’s double-check Bob #4:

v = 20 * 7.540 = 150.8 → divided by 36 = 4.1888... → squared = 17.546 → divided by 1.20 = 14.6217 → times 0.070 = 1.0235 → rounds to 1.024 N ✔

All calculations checked.

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Final Answer:

For the four bobs (using assumed experimental data), the centripetal forces are approximately:

- Bob #1 (shortest): 1.023 N
- Bob #2 (medium): 1.061 N
- Bob #3 (long): 1.035 N
- Bob #4 (longest): 1.024 N

*(Remember: Your actual answers will depend on your measured times and radii from the experiment.)*