A small coin is inside a bowl. The bowl is a surface of revolution of the curve y=100x4 m−3. This coin slides around the inside of the bowl at a constant height of y=0.01 m above the bottom of the bowl. What is its angular velocity in rad/s?

Question

please just say the method , i will calculate myself

Answer 1

At that height, you can evaluate the gradient of the normal(which is actually -2.5). then youu find the angle between the horizon and the normal at that height.Let the force acting from the surface along the normal as F.Decompose the force into F_x(parallel to the horizon) and F_y(perpendicular to the horizon).you will get
F_x = mg,
F_y = mv^2/r
then you can just solve for v by dividing equation 1 with equation 2

What you need to find: the radius at that height,the angle between the two lines.

Please tell me the numerical answer after u solve for the answer. Thanks!
(Remember it is the angular velocity.)

Answer 2

thank you a lot