Question

From the equation F = mv2/r, the tension in a 2 m length of string that whirls a 3 kg mass at 1 m/s in a horizontal circle is calculated to be 1.50 N. Calculate the tension for the following cases.
(a) twice the mass
(b) twice the speed
(c) twice the length of string (radial distance)
(d) twice mass, twice speed, and twice distance all at the same time

Answer 1

We know the formula: $F = \frac{mv^2}{r} = \frac{3 \cdot 1}{2} = 1.5 \, \text{N}$ .

(a) Twice the mass: $F = \frac{2mv^2}{r} = 2 \cdot \frac{mv^2}{r} = 2 \cdot 1.5 = 3 \, \text{N}$ .

Answer: 3 N.

(b) Twice speed: $F = \frac{m \cdot (2v)^2}{r} = 4 \cdot \frac{mv^2}{r} = 4 \cdot 1.5 = 6 \, \text{N}$ .

Answer: 6 N.

(c) Twice the length of string (radial distance): $F = \frac{mv^2}{2 \cdot r} = \frac{1}{2} \cdot \frac{mv^2}{r} = \frac{1}{2} \cdot 1.5 = 0.75 \, \text{N}$ .

Answer: 0.75 N.

(d) Twice mass, twice speed and twice distance all at the same time:

$$F = \frac {(2 m) \cdot (2 v) ^ {2}}{2 r} = \frac {2 \cdot 4}{2} \cdot \frac {m v ^ {2}}{r} = 4 \cdot \frac {m v ^ {2}}{r} = 4 \cdot 1. 5 = 6 \, \text{N}.$$

Answer: 6 N.