Here is my question:
A 200 N sphere 0.20 m\m in radius rolls, without slipping, 6.0 m down a ramp that is inclined 28 degrees with the horizontal. What is the angular speed of the sphere at the bottom of the hill if it starts from rest?
Do this with energy considerations:
changePE= rolling KE + translation KE
mg*6sin28 = 1/2 I w^2 + 1/2 mv^2
and of course, wr=v
mg*6sin28 = 1/2 I w^2 + 1/2 m (wr)^2
To find the angular speed of the sphere at the bottom of the hill, we can use the principle of conservation of energy.
The total initial potential energy of the sphere is given by the product of its mass (m), acceleration due to gravity (g), and the vertical distance it moves down the ramp, which is 6 sin(28°). This potential energy is entirely converted into the sum of the rolling kinetic energy and translational kinetic energy of the sphere at the bottom of the hill.
The rolling kinetic energy of the sphere depends on its moment of inertia (I) and its angular speed (ω). The translational kinetic energy depends on the mass (m) and linear velocity (v) of the sphere. We know that the linear velocity (v) and angular velocity (ω) are related by the equation v = rω, where r is the radius of the sphere.
So, plugging in the given values into the equation, we have:
mg * 6 sin(28°) = 1/2 I ω^2 + 1/2 m (rω)^2
Now, we need to find the moments of inertia for a solid sphere. The moment of inertia for a solid sphere rotating about its diameter is given by I = 2/5 * m * r^2.
Substituting this moment of inertia into the equation, we get:
mg * 6 sin(28°) = 1/2 (2/5 * m * r^2) ω^2 + 1/2 m (rω)^2
Simplifying further, we have:
mg * 6 sin(28°) = (1/5) m r^2 ω^2 + (1/2) m r^2 ω^2
Now let's solve for ω:
mg * 6 sin(28°) = (7/10) m r^2 ω^2
ω^2 = (10/7) * (g * 6 sin(28°)) / r^2
Taking the square root of both sides, we find:
ω = sqrt[(10/7) * (g * 6 sin(28°)) / r^2]
Now, substitute the known values (g = 9.8 m/s^2, r = 0.2 m), and calculate the angular speed (ω) using the above equation.