A ball of mass 21.7 g is attached to a cord of
length 0.61 m and rotates in a vertical circle.
What is the minimum speed the ball must
have at the top of the circle so the cord does
not become slack? The acceleration of gravity
is 9.8 m/s
.
Answer in units of m/s
Why did the ball have so many friends? Because it always stayed at the top of the circle and never let the cord become slack! To find the minimum speed the ball must have at the top, we can use the concept of centripetal force. To prevent the cord from becoming slack, the centripetal force must be greater than or equal to the tension in the cord.
At the top of the circle, the tension in the cord is equal to the weight of the ball (mg), where m is the mass of the ball and g is the acceleration due to gravity.
So, the centripetal force required at the top is also mg. Using the centripetal force equation (Fc = mv^2 / r), we can equate the centripetal force (mg) to the required force (mv^2 / r):
mg = mv^2 / r
Simplifying, we find:
g = v^2 / r
Now, we can solve for v:
v^2 = g * r
v = √(g * r)
Plugging in the given values (g = 9.8 m/s^2 and r = 0.61 m), we can calculate the minimum speed:
v = √(9.8 * 0.61)
v ≈ 3.07 m/s
So, the minimum speed the ball must have at the top of the circle is approximately 3.07 m/s.
To determine the minimum speed the ball must have at the top of the circle so that the cord does not become slack, we need to consider the forces acting on the ball at that point.
At the top of the circle, the tension in the cord is at its minimum. The forces acting on the ball are its weight (mg) and the tension in the cord (T). Since the cord does not become slack, the tension must be enough to balance the weight of the ball.
To find the tension, we need to equate the centripetal force (mv^2/r) to the sum of the weight and tension:
mv^2/r = mg + T
where:
m = mass of the ball = 21.7 g = 0.0217 kg
v = velocity of the ball at the top of the circle (what we need to find)
r = radius of the circle = length of the cord = 0.61 m
g = acceleration due to gravity = 9.8 m/s^2
Rearranging the equation, we get:
v^2 = (g + (T/m))r
Substituting the values, we have:
v^2 = (9.8 + (T/0.0217)) * 0.61
Since we are looking for the minimum speed, we can assume that the tension is at its minimum, which is zero:
v^2 = (9.8 + 0) * 0.61
v^2 = 5.978
Taking the square root of both sides to solve for v:
v = √(5.978)
v ≈ 2.447 m/s
Therefore, the minimum speed the ball must have at the top of the circle so that the cord does not become slack is approximately 2.447 m/s.
To find the minimum speed the ball must have at the top of the circle so that the cord does not become slack, we need to consider the forces acting on the ball at that position.
At the top of the circle, the gravitational force is directed downward and has a magnitude of m * g, where m is the mass of the ball (21.7 g = 0.0217 kg) and g is the acceleration due to gravity (9.8 m/s²). The tension force in the cord acts toward the center of the circle.
For the cord not to become slack, the tension force in the cord must be greater than or equal to the gravitational force acting downward.
We can use the concept of centripetal force to determine the tension in the cord at the top of the circle. The centripetal force required for an object with mass m rotating in a circle of radius r with speed v is given by:
Fc = (m * v²) / r
Since the cord length is given as r = 0.61 m, we can substitute the known values into the formula:
Fc = (0.0217 kg * v²) / 0.61 m
To prevent the cord from becoming slack, the tension force T must equal or exceed the gravitational force:
T ≥ m * g
Combining these inequalities, we get:
(m * v²) / r ≥ m * g
We can cancel out the masses (m) and simplify the inequality:
v² / r ≥ g
To find the minimum speed (v), we can rearrange the equation:
v ≥ √(g * r)
Plugging in the given values, we can calculate the minimum speed:
v ≥ √(9.8 m/s² * 0.61 m)
v ≥ √(5.978 m²/s²)
v ≥ 2.448 m/s
Therefore, the minimum speed the ball must have at the top of the circle so that the cord does not become slack is 2.448 m/s.