A ball on the end of a string is revolved at a uniform rate in a vertical circle of radius 72.0 cm, as shown in
Fig. 5–33. If its speed is .4 00 m s and its mass is 0.300 kg, calculate the tension in the string when the ball is
(a) at the top of its path, and (b) at the bottom of its path
To calculate the tension in the string when the ball is at the top and bottom of its path, we can use the concepts of centripetal force and gravitational force.
(a) Tension at the top of the path:
At the top of the path, the ball is moving in a circular motion, and the tension in the string provides the centripetal force required to keep the ball moving in a circular path. In addition to the centripetal force, the ball also experiences the force due to gravity acting downwards.
To calculate the tension at the top, we need to balance these two forces. The centripetal force is given by:
Fc = m * v^2 / r
Where:
m = mass of the ball = 0.300 kg
v = speed of the ball = 0.400 m/s
r = radius of the circle = 72.0 cm = 0.72 m
Substituting the values into the formula:
Fc = 0.300 kg * (0.400 m/s)^2 / 0.72 m
Calculate and simplify:
Fc = 0.300 kg * 0.160 m^2/s^2 / 0.72 m
Fc = 0.0667 kg * m/s^2
Therefore, the centripetal force at the top of the path is 0.0667 N.
At the top of the path, the tension force and gravitational force add up because they are in the same direction, i.e., downwards. So the total force at the top is the sum of these two forces:
Ftotal = Fc + Fg
Where:
Fg = force due to gravity = m * g
m = mass of the ball = 0.300 kg
g = acceleration due to gravity = 9.8 m/s^2 (approx)
Substituting the values:
Fg = 0.300 kg * 9.8 m/s^2
Calculate:
Fg = 2.94 N
Therefore, the total force at the top is:
Ftotal = 0.0667 N + 2.94 N
Calculate:
Ftotal = 3.01 N
Hence, the tension in the string at the top of the path is approximately 3.01 N.
(b) Tension at the bottom of the path:
At the bottom of the path, the tension in the string needs to counteract both the centripetal force and the force due to gravity. However, this time they act in opposite directions.
The centripetal force is still given by:
Fc = m * v^2 / r
Substituting the given values:
Fc = 0.300 kg * (0.400 m/s)^2 / 0.72 m
Calculate:
Fc = 0.0667 N
The force due to gravity is still given by:
Fg = m * g
Substituting the values:
Fg = 0.300 kg * 9.8 m/s^2
Calculate:
Fg = 2.94 N
At the bottom of the path, the tension and gravitational force subtract because they are in opposite directions. So the total force at the bottom is the difference between these two forces:
Ftotal = Fc - Fg
Substituting the calculated values:
Ftotal = 0.0667 N - 2.94 N
Calculate:
Ftotal = -2.87 N
Note that the negative sign indicates that the force is acting in the opposite direction to the chosen positive direction. In this case, it suggests that the tension is not sufficient to counteract the gravitational force.
Hence, the tension in the string at the bottom of the path is approximately -2.87 N.