an annular ring starts from rest and rolls without sliding down a hill from a height of 22.5m. it has an outer radius from 1.0 m and an inner radius of 0.50m. if the moment of inertia of an annular ring in I=1/2M(ro^2 + ri^2), what is th elinear velocity of the ring when it reaches the bottom of the hill.
To find the linear velocity of the annular ring when it reaches the bottom of the hill, we can use the principle of conservation of energy.
1. First, let's calculate the potential energy at the top of the hill, which will be converted into kinetic energy at the bottom.
Potential Energy (PE) = mass (m) * gravitational acceleration (g) * height (h)
PE = mgh
In this case, the height is given as 22.5 meters, and the gravitational acceleration is approximately 9.8 m/s^2.
2. Next, we need to relate the moment of inertia (I) of the annular ring to its rotational kinetic energy.
Rotational Kinetic Energy (KE_rot) = (1/2) I ω^2
Where ω is the angular velocity.
The moment of inertia of the annular ring is given by I = (1/2)M(r_outer^2 + r_inner^2)
Here, the outer radius (r_outer) is 1.0 m, and the inner radius (r_inner) is 0.50 m.
3. At the bottom of the hill, when the annular ring reaches its maximum linear velocity (v), all the potential energy is converted into kinetic energy.
Therefore, PE = KE_rot + KE_linear
Since the annular ring rolls without sliding, the linear velocity is related to the angular velocity by v = ω (r_outer - r_inner)
Now we can put all these pieces together to solve for the linear velocity.
1. Calculate the potential energy:
PE = m * g * h
PE = m * 9.8 m/s^2 * 22.5 m
2. Calculate the moment of inertia:
I = (1/2)M(r_outer^2 + r_inner^2)
I = (1/2)M(1.0 m^2 + 0.50 m^2)
3. Equate the potential energy to the sum of rotational and linear kinetic energy:
PE = KE_rot + KE_linear
Substitute the relevant equations:
m * 9.8 m/s^2 * 22.5 m = (1/2) I ω^2 + (1/2) m v^2
Since v = ω (r_outer - r_inner), we have:
m * 9.8 m/s^2 * 22.5 m = (1/2) I ω^2 + (1/2) m (ω (r_outer - r_inner))^2
4. Simplify the equation and solve for v:
Rearrange the equation to solve for v:
v^2 = 2 * m * 9.8 m/s^2 * 22.5 m / (m + I / (r_outer - r_inner)^2)
Take the square root to find v:
v = √(2 * m * 9.8 m/s^2 * 22.5 m / (m + I / (r_outer - r_inner)^2))
Now you can substitute the given values (mass, outer radius, inner radius) into the equation to find the linear velocity (v) of the ring when it reaches the bottom of the hill.