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?

please just say the method , i will calculate myself

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.)

thank you a lot

To calculate the angular velocity of the coin sliding inside the bowl, you need to determine the radius of the bowl at the height of the coin.

The equation of the curve for the bowl is given as y = 100x^4 m^-3. We can rearrange this equation to solve for x:
x = (y/100)^(1/4)

The height at which the coin slides is y = 0.01 m. Plugging this value into the equation, we can find the corresponding x-coordinate:
x = (0.01/100)^(1/4)
x = 0.1^(1/4)
x = 0.3162

Now, we can use the x-coordinate to calculate the radius of the bowl at the height of the coin. Since the bowl is a surface of revolution, the radius at the given height is the x-coordinate itself:
r = 0.3162 m

The angular velocity (ω) of the coin can be calculated using the formula:
ω = v / r,

where v is the linear velocity of the coin. Since the coin is sliding at a constant height of 0.01 m, the linear velocity is also constant. Therefore, you need to determine the linear velocity at y = 0.01 m. To do this, you can differentiate the equation y = 100x^4 m^-3 with respect to time, assuming that the y-coordinate changes with time:

dy/dt = d(100x^4 m^-3)/dt
= 400x^3 * dx/dt
= 400x^3 * v,

where dy/dt is the rate of change of the y-coordinate with respect to time (which is equal to v), dx/dt is the rate of change of the x-coordinate with respect to time, and v is the linear velocity.

Since the coin is sliding around at a constant height, dy/dt is equal to zero. Therefore, we have:

0 = 400x^3 * v.

Solving for v:

v = 0.

Since the linear velocity is 0, the angular velocity of the coin is also 0 rad/s.

Therefore, the angular velocity of the coin sliding around the inside of the bowl at a constant height of y = 0.01 m is 0 rad/s.