Question

an owl m=4.72kg slides down a semi-circular ramp r=7.5m. the owl starts at the top ramp at rest. when the owl is at an angle 31.9. what is the normal force exerted by the owl.

Answer 1

The situation can be visualized by

The total energy of the block at the top is

And the total energy at an angle θ is

The conservation of energy gives us

The centripetal force is given by

and for angle θ

The free-body diagram for the owl at the angle θ is

The force in the radial direction is given by

This force is responsible for the circular motion. So, this force must be equal to the centripetal force. i.e

Now putting the values in the equation

So, the normal force exerted by the circular surface on the owl is 25.3 N.

We know the normal forces come in pairs with equal magnitude and opposite in direction.

So, the normal force exerted by the owl is 25.3 N.