A man is swirling a stone tied to one end of a horizontal string The stone is in uniform circular motion and makes 20 revolutions in 4 seconds

Question

a. What is the frequency of the motion?

b. What is its period?

c. What is the speed of the stone if the length of the string that holds the stone is 0.8m?

d. What is the centripetal force acting on the stone to keep it in uniform circular motion if its mass is 30 g?

GPT-4o mini · Answer

To solve the problem, we will use the definitions and formulas related to circular motion.

Frequency ( $f$ ) is defined as the number of revolutions per second.

Given:

$f = \frac{Number of revolutions}{Time (in seconds)} = \frac{20 revolutions}{4 seconds} = 5 Hz$

The period ( $T$ ) is the time taken for one complete revolution, and it is the reciprocal of frequency.

$T = \frac{1}{f} = \frac{1}{5 Hz} = 0.2 seconds$

The speed ( $v$ ) of the stone in uniform circular motion can be calculated using the formula:

$v = \frac{2 π r}{T}$

where $r$ is the radius (which is the length of the string) and $T$ is the period.

Given:

Substituting the values:

$v = \frac{2 π (0.8 m)}{0.2 s} = \frac{1.6 π m}{0.2 s} = 8 π m/s \approx 25.13 m/s$

First, we need to convert the mass into kilograms:

$m = 30 g = 0.03 kg$

The centripetal force ( $F_{c}$ ) required to keep an object in circular motion is given by the formula:

$F_{c} = \frac{m v^{2}}{r}$

Substituting the values:

Now calculating $v^{2}$ :

$v^{2} \approx (25.13)^{2} \approx 631.63 m^{2} / s^{2}$

Now substitute into the centripetal force formula:

$F_{c} = \frac{0.03 kg \times 631.63 m^{2} / s^{2}}{0.8 m} \approx \frac{18.949 {kg m}^{2} / s^{2}}{0.8} \approx 23.686 N$