When a 75kg goes over the top of a hemispherical bump at a speed of 18m/s their contact force with the ground is just 250N. What is the radius of the bump?
To solve this problem, we need to apply the principles of motion and forces. We can use the concept of centripetal force to calculate the radius of the bump.
Centripetal force is the force exerted on an object moving in a circular path. It is given by the equation:
F = (m * v^2) / r
Where:
F is the centripetal force
m is the mass of the object
v is the speed of the object
r is the radius of the circular path
In this case, the vertical force exerted by the object when it goes over the bump is equal to its weight minus the contact force with the ground:
Vertical force = weight - contact force
Vertical force = m * g - contact force
In the problem, the contact force is given as 250N, and the mass is 75kg. The acceleration due to gravity (g) is approximately 9.8 m/s^2.
Vertical force = (75 kg * 9.8 m/s^2) - 250 N
Now, let's calculate the vertical force:
Vertical force = 735 N - 250 N
Vertical force = 485 N
Therefore, the vertical force acting on the object is 485N.
Now, we can equate the centripetal force to the vertical force:
(m * v^2) / r = vertical force
(75 kg * (18 m/s)^2) / r = 485 N
To find the radius of the bump, rearrange the equation:
r = (75 kg * (18 m/s)^2) / 485 N
Now, calculate the radius:
r = 1.7 m
Therefore, the radius of the bump is approximately 1.7 meters.