A 50-kg person starts from rest at the top of a large frictionless spherical surface (radius 6 m) and slides into the water below. What is the magnitude of the normal force she experiences halfway along the spherical surface (at theta = 45 degrees as shown)? At this point, she is moving with a tangential velocity of 4.0 m/s.
To find the magnitude of the normal force experienced by the person halfway along the spherical surface, we can use the principles of circular motion.
First, let's find the acceleration of the person at that point. The only force acting on the person is gravity, which is pointing vertically downwards. Since the surface is frictionless, there is no other force acting on the person along the surface.
At any point along the surface, the weight of the person provides the centripetal force required for circular motion:
F_c = m * a_c
Where F_c is the centripetal force, m is the mass of the person, and a_c is the centripetal acceleration.
The centripetal force is provided by the component of the person's weight along the surface:
F_c = mg sin(theta)
Where g is the acceleration due to gravity and theta is the angle measured from the top of the sphere. At theta = 45 degrees, the centripetal force becomes:
F_c = mg sin(45)
Next, we can find the centripetal acceleration using the equation:
a_c = v^2 / r
Where v is the tangential velocity of the person and r is the radius of the spherical surface.
At the halfway point along the spherical surface, the radius is half of the total radius:
r = 6 m / 2 = 3 m
So, the centripetal acceleration becomes:
a_c = (4.0 m/s)^2 / 3 m
Now, we can plug in the values to find the centripetal force:
F_c = (50 kg) * (9.8 m/s^2) * sin(45) = 343 N
Since the normal force is equal to the centripetal force (as there is no horizontal force), the magnitude of the normal force experienced by the person halfway along the spherical surface is 343 N.