A curious child finds a rope hanging vertically from the ceiling of a large storage hangar. The child grabs the rope and starts running in a circle. The length of the rope is 13.0m. When the child runs in a circle of radius 7.0m, the child is about to lose contact with the floor. How fast is the child running at that time?
v= ?
I know I have to apply the centripetal acceleration, but I don't know where to go about with the distance
Is the final answer 0.82 m/s
Nevermind, I got 6.6m/s
To solve this problem, we need to apply the centripetal acceleration formula and determine the speed of the child when they are about to lose contact with the floor.
The centripetal acceleration formula is given by:
a = v^2 / r,
where a is the centripetal acceleration, v is the velocity, and r is the radius of the circular path.
In this problem, the radius (r) is given as 7.0m. We need to find the velocity (v).
To find the distance (d) the child covers, you can calculate the circumference of the circle using the formula:
d = 2πr,
where d is the distance and r is the radius.
Given that the length of the rope is 13.0m, we can equate the distance to the length of the rope:
d = 2πr = 13.0m.
Now, we can solve this equation to find the value of r:
2πr = 13.0m,
r = 13.0m / (2π).
Plugging in this value of r into the centripetal acceleration formula:
a = v^2 / r,
v^2 = a * r,
v = (a * r)^0.5.
In this case, the centripetal acceleration is given by gravity (9.8 m/s^2) since the child is about to lose contact with the floor.
Now you can substitute the values and calculate the velocity (v) when the child is about to lose contact with the floor.
the horizontal component of the tension in the rope supplies the centripetal force that keeps the child moving in the circle
... m v^2 / r = T * (7 / 13)
the vertical component of the tension is equal to the child's weight
... m g = T [√(13^2 - 7^2) / 13]