A certain string can withstand a maximum tension of 41 N without breaking. A child ties a 0.35 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?
a)at a random point on the path
b)cannot be determined
c)at the lowest point on the path
d)at the highest point on the path
(b) What is the speed of the stone as the string breaks?
m/s
Abbreviating the subject as "p" is not helpful to teachers looking for questions in their area of expertise.
The string will break where the stress is highest. At the bottom of the vertical loop, the string tension T must balance both weight and centripetal forces.
T = M [(V^2/R) + g]
At the top,
T = M [(V^2/R) - g]
(a) The stone will break the string at the lowest point on its path (c) because this is where it experiences the maximum tension. At the lowest point, the weight of the stone is added to the tension force and the string has to withstand the full force of the stone's weight plus the centrifugal force.
(b) To determine the speed of the stone as the string breaks, we can use the concept of centripetal force:
The tension in the string is responsible for providing the centripetal force to keep the stone moving in a circular path. At the lowest point, the tension force and the weight of the stone both contribute to the centripetal force.
The centripetal force is given by:
T = m * g + m * v^2 / r
where T is the tension force, m is the mass of the stone, g is the acceleration due to gravity, v is the speed of the stone, and r is the radius of the circle.
The weight of the stone is given by:
m * g = 0.35 kg * 9.8 m/s^2 = 3.43 N
The tension force should equal or exceed the maximum tension of the string, which is 41 N:
T = m * g + m * v^2 / r >= 41 N
Substituting the known values, we have:
3.43 N + 0.35 kg * v^2 / 0.91 m >= 41 N
Rearranging the equation:
0.35 kg * v^2 / 0.91 m >= 41 N - 3.43 N
0.35 kg * v^2 / 0.91 m >= 37.57 N
Multiplying both sides by 0.91 m:
0.35 kg * v^2 >= 0.91 m * 37.57 N
Dividing by 0.35 kg:
v^2 >= (0.91 m * 37.57 N) / 0.35 kg
v^2 >= 244.17 m^2/s^2
Taking the square root of both sides:
v >= sqrt(244.17) m/s
v >= 15.63 m/s
Therefore, the speed of the stone as the string breaks is at least 15.63 m/s.
To determine where the stone is on its path when the string breaks, we need to identify the specific condition that would cause the string to break. In this case, the string can withstand a maximum tension of 41 N before breaking.
As the stone moves in a vertical circle, its weight generates two forces: tension in the string and gravitational force. When the stone reaches the highest point on the path, the string is under the most tension due to the combination of the stone's weight and the upward acceleration required to keep it moving in a circle. Therefore, the tension in the string would be at its maximum at the highest point on the path.
Given that the string can withstand a maximum tension of 41 N, we can conclude that the string will break at the highest point on the stone's path.
Therefore, the correct answer to (a) is: d) at the highest point on the path.
Moving on to (b), we need to find the speed of the stone when the string breaks. To do this, we can use the concept of centripetal force, which is the force that keeps an object moving in a circular path.
At any point on the path, the centripetal force required is provided by tension in the string:
Centripetal Force = Tension in string
For the stone to move in a vertical circle, the centripetal force must equal the net force acting on the stone. The net force is the difference between weight and tension:
Net Force = Weight - Tension in string
At the point when the string breaks, the tension becomes zero. Therefore, we can write:
Net Force = Weight - 0
Now, we can set up the equation using the given values:
m * g = m * v^2 / r
where,
m = mass of the stone (0.35 kg),
g = acceleration due to gravity (9.8 m/s^2),
v = speed of the stone, and
r = radius of the circular path (0.91 m).
Simplifying the equation:
0.35 kg * 9.8 m/s^2 = 0.35 kg * v^2 / 0.91 m
v^2 = 0.35 kg * 9.8 m/s^2 * 0.91 m / 0.35 kg
v^2 = 9.8 m^2/s^2 * 0.91 m
v^2 = 8.938 m^2/s^2
v = sqrt(8.938 m^2/s^2)
v ≈ 2.99 m/s
Therefore, the speed of the stone as the string breaks is approximately 2.99 m/s.