a man swings his child in a circle of radius .75 meters. if the mass of the child is 25 kilograms and the child makes one revolution in 1.5 seconds, what is the magnitude and direction of the force that must be exerted by the man on the child?
Ac = w^2 r
w = 2 pi radians/1.5 seconds = 4.19 radians/second
Ac = 4.19^2 * .75 = 13.2 m/s^2
Fin = m Ac = 25*13.2 = 329 Newtons
Fup = m g = 25 * 9.81 = 245 Newtons
magnitude of force = sqrt(329^2+245^2)
tangent of direction up from horizontal = 245/329
To find the magnitude and direction of the force exerted by the man on the child, we can use the concept of centripetal force.
Centripetal force is the force that keeps an object moving in a circle, directed toward the center of the circle. In this case, the force exerted by the man on the child acts as the centripetal force.
The formula for centripetal force is:
F = (m * v^2) / r
Where:
F is the centripetal force
m is the mass of the child
v is the velocity of the child
r is the radius of the circle
Given:
m = 25 kg
r = 0.75 m
v = (2 * π * r) / t (since the child makes one revolution in 1.5 seconds)
Let's substitute the given values into the formula to calculate the force:
v = (2 * π * 0.75) / 1.5
v ≈ 3.14 m/s
Now, substitute the values of m, v, and r into the formula:
F = (25 * (3.14^2)) / 0.75
F ≈ 329.71 N
So, the magnitude of the force that must be exerted by the man on the child is approximately 329.71 Newtons. The direction of the force is towards the center of the circle, which is the force that keeps the child moving in the circular path.