A ball of mass m 0.285 kg swings in a vertical circular path on a string L = 0.810 m long as in the figure below.
a) If its speed is 5.25 m/s at the top of the circle, what is the tension in the string there? 6.89N (is the answer)
b)If the string breaks when its tension exceeds 20.5 N, what is the maximum speed the ball can have at the bottom before that happens?
At the top:
mv^2/r=Tension+mg
tension= -mg+mv^2/r
= .285(-9.8+5.25^2/.81)
To solve this problem, we need to use the concept of centripetal force and Newton's second law.
a) At the top of the circular path, when the ball is moving with a speed of 5.25 m/s, the tension in the string is equal to the centripetal force required to keep the ball moving in a circular path.
We can start by finding the centripetal force using the formula:
F_c = (m * v^2) / r
Where:
F_c is the centripetal force,
m is the mass of the ball,
v is the speed of the ball,
and r is the radius of the circular path.
In this case, the radius of the circular path is equal to the length of the string, L, which is given as 0.810 m.
So, substituting the given values, we have:
F_c = (0.285 kg * (5.25 m/s)^2) / 0.810 m
Calculating this expression, we find:
F_c ≈ 6.89 N
Therefore, the tension in the string at the top is approximately 6.89 N.
b) To find the maximum speed the ball can have at the bottom before the string breaks, we need to find the maximum centripetal force that the string can withstand before breaking.
Given that the maximum tension the string can handle is 20.5 N, we can set the centripetal force equal to this maximum tension:
F_c = 20.5 N
Using the same formula as before and solving for v, we get:
20.5 N = (0.285 kg * v^2) / 0.810 m
Rearranging this equation and solving for v, we find:
v^2 = (20.5 N * 0.810 m) / 0.285 kg
v^2 ≈ 57.518 m^2/s^2
Taking the square root of both sides, we get:
v ≈ √(57.518 m^2/s^2)
v ≈ 7.59 m/s
Therefore, the maximum speed the ball can have at the bottom before the string breaks is approximately 7.59 m/s.
To answer these questions, we can use the principles of centripetal force and gravitational force.
a) To find the tension in the string at the top of the circle, we need to consider the forces acting on the ball. At the top, the tension in the string and the gravitational force are both acting towards the center of the circular path.
To find the tension, we can use the equation for centripetal force:
F_c = (m * v^2) / r
where F_c is the centripetal force, m is the mass of the ball, v is its velocity, and r is the radius (L) of the circular path.
In this case, the centripetal force is provided by the tension in the string. So we can write:
Tension = (m * v^2) / r
Substituting the given values:
Tension = (0.285 kg * (5.25 m/s)^2) / 0.810 m
Tension = 6.89 N
Therefore, the tension in the string at the top of the circle is indeed 6.89 N.
b) To find the maximum speed the ball can have at the bottom before the string breaks, we need to consider the tension in the string at the bottom of the circle. At the bottom, the tension and gravitational force are acting in the same direction, adding up to create a greater tension.
The maximum tension the string can handle before breaking is given as 20.5 N. So we can write:
Tension = mg + m * v^2 / r
where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), v is the velocity of the ball at the bottom, and r is the radius of the circular path.
Rearranging the equation, we can solve for v:
v^2 = (Tension - mg) * r / m
v = sqrt((Tension - mg) * r / m)
Substituting the given values:
v = sqrt((20.5 N - (0.285 kg * 9.8 m/s^2)) * 0.810 m / 0.285 kg)
v = 7.92 m/s
Therefore, the maximum speed the ball can have at the bottom before the string breaks is approximately 7.92 m/s.