0.78 kg object attached to a 4.47 m string is moving in a circular motion on a vertical plane with 50 m/s uniform speed. What is the tension of the string at the lowest point of the circle?

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The problem gives a mass m = 0.78 kg, a circular path with radius r = 4.47 m, and a constant speed v = 50 m/s in a vertical plane. We are asked for the tension in the string at the lowest point of the circle.

First, recall the forces on the mass at the bottom of the swing. At the lowest point, the tension T in the string acts upward, and the weight mg acts downward. The net centripetal force toward the center (upward, toward the circle’s center) is provided by the combination of tension and weight:
T − mg = m v^2 / r.
Rearranging gives T = mg + m v^2 / r. This is a standard result for circular motion in a vertical plane at the bottom of the path.

Compute the components:
- m g is the weight: g is typically 9.8 m/s^2, so mg = 0.78 × 9.8 ≈ 7.644 N. For quick checks, you can approximate g as 9.8 or 9.81; the result will be close.
- m v^2 / r is the centripetal term: v^2 = (50 m/s)^2 = 2500 (m^2/s^2). Then m v^2 / r = 0.78 × 2500 / 4.47.

Do the division: 0.78 × 2500 = 1950. Now divide by 4.47: 1950 / 4.47 ≈ 436.3. You can verify by rough estimation: 4.5 × 436 ≈ 1962, so slightly less than 436 would give about 1950, which aligns with 436.3.

Add the two contributions for the tension: T ≈ 7.644 + 436.3 ≈ 443.9 N. Rounding to the nearest whole number gives about 444 N.

Key checks and possible pitfalls:
- The formula T = mg + m v^2 / r is specific to the lowest point, where tension must support weight and provide the centripetal acceleration upward.
- If one mistakenly used T = m v^2 / r alone, the result would be underestimated by neglecting the weight mg.
- If the speed were different or the radius changed, both terms would scale differently, so always confirm which point in the path you are analyzing and which forces contribute to the centripetal requirement.

Summary of the calculation steps:
1) Compute mg = 0.78 × 9.8 ≈ 7.644 N.
2) Compute m v^2 / r = (0.78 × 50^2) / 4.47 ≈ 1950 / 4.47 ≈ 436.3 N.
3) Sum to obtain T = mg + m v^2 / r ≈ 7.644 + 436.3 ≈ 443.9 N.

MUF0121 Physics Unit 1 - Semester 2, 2025 8.15 Vertical circular motion

0.78 kg object attached to a 4.47 m string is moving in a circular motion on a vertical plane with 50 m/s uniform speed. What is the tension of the string at the lowest point of the circle?

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