A certain string can withstand a maximum tension of 40 N without breaking. A child ties a 0.37 kg stone to one end and, holding the other end, whirls the stone in a vertical circle of radius 0.91 m, slowly increasing the speed until the string breaks.
(a) Where is the stone on its path when the string breaks?
(b) What is the speed of the stone as the string breaks?
To determine the point where the string breaks and the speed of the stone at that moment, we can use the centripetal force equation and apply it to the situation described.
(a) To find where the stone is on its path when the string breaks, we need to determine when the tension in the string reaches its maximum allowable tension of 40 N.
The tension in the string can be calculated using the following equation:
Tension = (mass of stone) * (velocity^2) / (radius)
where:
mass of stone = 0.37 kg (given)
velocity = speed of the stone
radius = 0.91 m (given)
We can rearrange the equation to solve for the velocity:
velocity^2 = (Tension * radius) / (mass of stone)
Substituting the given values:
velocity^2 = (40 N * 0.91 m) / 0.37 kg
velocity^2 = 98.38 N*m/kg
Taking the square root of both sides:
velocity ≈ 9.92 m/s
(b) Now that we have the speed of the stone as the string breaks, we can find the position of the stone on its path.
At the moment the string breaks, the tension is at its maximum. This occurs when the stone is at the bottom of its circular path, where the tension is greatest due to the stone's weight.
Therefore, the stone is at the lowest point on its path when the string breaks.
In summary:
(a) The stone is at the lowest point of its circular path when the string breaks.
(b) The speed of the stone at the moment the string breaks is approximately 9.92 m/s.
To determine where the stone is on its path when the string breaks (part a), we need to consider the tension in the string and the net force acting on the stone. We know that the maximum tension the string can withstand without breaking is 40 N.
In this scenario, the tension in the string is caused by two forces:
1. The tension due to the weight of the stone: Tension_weight = m * g, where m is the mass of the stone and g is the acceleration due to gravity (approximately 9.8 m/s²).
2. The tension due to the acceleration of the stone as it moves in a circle: Tension_centripetal = m * (v² / r), where v is the velocity of the stone and r is the radius of the circle.
Since the child is increasing the speed of the stone until the string breaks, we can assume that the tension in the string reaches its maximum value of 40 N when the string breaks. Therefore, both the weight and the centripetal tension must sum up to equal 40 N:
Tension_weight + Tension_centripetal = 40 N
Now, let's solve part a by finding the position of the stone on its path when the string breaks:
When the stone is at the lowest point of its path, the net force acting on it is the difference between the tension (T) and the weight (mg):
Net_force = T - mg
At the lowest point of the circular path, the net force is directed upwards and is equal to the centripetal force required to keep the stone moving in a circle:
Net_force = T - mg = m * (v² / r)
Since we know the tension T is equal to 40 N, we can rearrange the equation:
40 N - mg = m * (v² / r)
Now, let's calculate the position of the stone (part a) and the speed of the stone (part b):
Step 1: Calculate the position of the stone:
- Plug in the values: m = 0.37 kg, g = 9.8 m/s², r = 0.91 m
- Solve the equation for v²: 40 N - (0.37 kg * 9.8 m/s²) = 0.37 kg * (v² / 0.91 m)
- Simplify further: 40 N - 3.6266 N = v² / 0.91 m
- Solve for v²: 36.3734 N = v² / 0.91 m
- Multiply both sides by 0.91 m: v² = 36.3734 N * 0.91 m
- Calculate: v² = 33.15 Nm
- Take the square root of both sides to find v: v ≈ √33.15 Nm
Step 2: Calculate the speed of the stone:
- Substitute the value of v into the equation: v ≈ √33.15 Nm
By following these steps, you can calculate the position of the stone on its path when the string breaks (part a) and the speed of the stone (part b).