Q.A stone of mass $0.25\ \text{kg}$ tied to the end of a string is whirled round in a circle of radius $1.5\ \text{m}$ with a speed of $40\ \text{rev./min}$ in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of $200\ \text{N}$?

The key idea here is that the tension in the string provides the necessary centripetal force to keep the stone moving in a circular path.

1. First, convert the angular speed from revolutions per minute to radians per second: $\omega = 40\ \text{rev./min} = 40 \times \frac{2\pi\ \text{rad}}{60\ \text{s}} = \frac{4\pi}{3}\ \text{rad/s}$
2. The tension $T$ in the string is equal to the centripetal force $F_c = mr\omega^2$: $T = (0.25\ \text{kg})(1.5\ \text{m})\left(\frac{4\pi}{3}\ \text{rad/s}\right)^2 = 0.25 \times 1.5 \times \frac{16\pi^2}{9} = \frac{2\pi^2}{3}\ \text{N} \approx 6.58\ \text{N}$
3. For the maximum speed, the tension $T_{max}$ provides the maximum centripetal force. We use $T_{max} = \frac{mv_{max}^2}{r}$: …

When an object moves in a circular path, it constantly changes direction. This change in direction implies an acceleration, even if the speed remains constant. This acceleration is directed towards the center of the circle and is called centripetal acceleration. According to Newton's second law, an acceleration must be caused by a net force. This force, also directed towards the center of the circle, is known as the centripetal force.

In this problem, a stone is whirled in a horizontal circle. The string connecting the stone to the center provides the necessary centripetal force. The force exerted by the string is its tension. Therefore, the tension in the string is precisely the centripetal force required to keep the stone moving in its circular path. We assume the horizontal plane is perfectly flat, so gravity acts vertically downwards and is balanced by some other vertical force (e.g., a slight upward component from the string if it sags, or a normal force if it's on a surface), and does not contribute to the horizontal centripetal force.

The magnitude of the centripetal force ($F_c$) is given by the formula:

Let's break down the problem into steps.

1. Identify Given Quantities and Convert Units:

   We are given:

   • Mass of the stone, $m = 0.25\ \text{kg}$
   • Radius of the circle, $r = 1.5\ \text{m}$
   • Speed of the stone, $40\ \text{rev./min}$

   The speed needs to be converted into standard units of linear speed (meters per second, m/s).

   One revolution corresponds to a distance of $2\pi r$.

   One minute is $60$ seconds.

   First, let's convert revolutions per minute to revolutions per second:

   $40\ \text{rev./min} = \frac{40\ \text{rev}}{1\ \text{min}} \times \frac{1\ \text{min}}{60\ \text{s}} = \frac{40}{60}\ \text{rev./s} = \frac{2}{3}\ \text{rev./s}$

   Now, calculate the linear speed $v$:

   $v = (\text{revolutions per second}) \times (\text{circumference per revolution})$

   $v = \frac{2}{3}\ \text{rev./s} \times (2\pi r)\ \text{m/rev.}$

   $v = \frac{2}{3} \times 2\pi \times 1.5\ \text{m/s}$

   $v = \frac{2}{3} \times 2\pi \times \frac{3}{2}\ \text{m/s}$

   $v = 2\pi\ \text{m/s}$

   Using $\pi \approx 3.14159$:

   $v \approx 2 \times 3.14159\ \text{m/s} \approx 6.283\ \text{m/s}$

2. Calculate the Tension in the String ($T_1$):

   The tension in the string provides the centripetal force.

   $T_1 = F_c = \frac{mv^2}{r}$

   Substitute the values:

   $T_1 = \frac{0.25\ \text{kg} \times (2\pi\ \text{m/s})^2}{1.5\ \text{m}}$

   $T_1 = \frac{0.25 \times 4\pi^2}{1.5}\ \text{N}$

   $T_1 = \frac{\pi^2}{1.5}\ \text{N}$

   $T_1 = \frac{2\pi^2}{3}\ \text{N}$ …

Concept: Centripetal Force Supplied by String Tension

Step 1: Convert the given rate to angular speed

Step 2: Tension = centripetal force

The string supplies the centripetal force $F_c=mr\omega^2$:

Step 3: Evaluate $T$

Step 4: Find the maximum speed from the maximum tension

Set the centripetal-force requirement equal to $T_{\max}=200\ \text{N}$: …