A ball is tied to one end of a string. The other end of the string is held fixed. The ball is set moving around a vertical circle without friction, and with speed vi = (Rg)1/2. At what angle from the bottom should the string be cut so that the ball will then travel through the center of the circle on its way to the ground?
I am wondering: what was the initial position of the ball.
After pushing a 22.2 kg kid and tricycle over 10.3 m of smooth, level sidewalk from rest, the "combo" is moving at 3.0 m/s. Find the combo's kinetic energy
To find the angle at which the string should be cut, we need to consider the forces acting on the ball at that point.
1. First, let's draw a free-body diagram of the ball when it is at the bottom of the circular motion (angle = 0°). The forces acting on the ball at this point are its weight (mg) and the tension in the string (T).
Tension (T)
|
|
__ |
/ |
/ |
m/theta |
/ |
/________|
Ground
2. At the bottom of the circular motion, the tension in the string provides the centripetal force required for the ball to move in a circular path. Thus, we can write the equation:
T = mv^2 / R
where m is the mass of the ball, v is the velocity of the ball at the bottom (given as vi = (Rg)^(1/2)), and R is the radius of the circle.
3. Now, let's consider the forces acting on the ball when the string is cut (angle = θ). At this point, only the weight of the ball (mg) acts on it.
Tension (zero)
|
|
__ |
/ |
/ |
m/theta |
/ |
/________|
Ground
4. Since the string is cut, the ball will move in a straight line downward. So, the force of gravity will act as the net force:
F_net = mg
5. We can equate the tension in the string at the bottom with the net force acting on the ball when the string is cut, as they should be equal:
T = F_net
mv^2 / R = mg
6. Rearranging the equation, we can solve for the velocity v at the bottom:
v^2 = Rg
7. Substituting this value of v^2 back into equation (5):
m(Rg) / R = mg
8. Cancelling out the mass (m) and simplifying:
Rg / R = g
9. Cancelling out the g on both sides of the equation:
R / R = 1
10. This equation states that the radius of the circle cancels out, indicating that the angle θ at which the string is cut does not depend on the radius of the circle. Therefore, the angle θ at which the string should be cut is the same for any circular path.
11. The angle θ at which the string should be cut is 90° from the bottom, or straight down.
To determine at what angle from the bottom the string should be cut, we need to consider the forces acting on the ball as it moves in a vertical circle.
First, let's analyze the forces when the ball is at the bottom of the circle (θ = 0°). At this point, the tension in the string is directed vertically upward to balance the weight of the ball, which acts vertically downward. The net force on the ball is the centripetal force required to keep it moving in a circular path.
Next, let's consider the forces when the ball is at the desired angle θ. At this point, we want the ball to travel through the center of the circle on its way to the ground. To achieve this, the net force on the ball must be directed vertically downward.
To find the angle at which the string should be cut, we need to determine when the net force changes direction from upward to downward. At this transitioning point, the net force must pass through the center of the circle.
Using the concept of centripetal force, we can set up an equation to equate the tension force T (upward) at the bottom of the circle to the gravitational force mg (downward) acting at the desired angle θ:
T = mg
Since the velocity of the ball at the bottom of the circle is given by vi = (Rg)^(1/2), we can also equate the centripetal force required to maintain circular motion with the weight of the ball:
mv^2/R = mg
Notice that the mass 'm' cancels out in both equations, allowing us to eliminate it. Here, 'v' represents the velocity of the ball at an angle θ.
Now, we can find an expression for 'v' in terms of θ. The velocity 'v' can be expressed in terms of the initial velocity at the bottom of the circle (vi) using conservation of energy:
mgR (1 - cosθ) = (1/2)mv^2
Since vi^2 = 2gR, we can rewrite the equation as:
mgR(1 - cosθ) = mv^2/2 = m(vi^2)/2 = m(gR)/2
Simplifying the equation, we have:
(1 - cosθ) = 1/2
Solving for cosθ, we find:
cosθ = 1/2
Therefore, the angle θ at which the string should be cut is equal to the inverse cosine of 1/2, which is 60°.
Hence, the string should be cut at an angle of 60° from the bottom to ensure that the ball travels through the center of the circle on its way to the ground.