Asolid spherical ball starts from rest at height h at the top of a hill and begins rolling without slipping down the hill.

(a) what is the speed of the ball when it is halfway down the hill?

(b) at the halfway point (height h/2), the hill becomes smooth, so there is no longer any friction between the hill and the sphere. what is the speed of the ball at the bottom of the hill?

a) √((5/7)*g*h)     b) √((12/7)*g*h)

Explanation:

a) Applying mechanical energy conservation, we can prove that the sum of  the change in the gravitational energy and  the change in the kinetic energy is equal to 0, as no non-conservative force does any work.

Friction is static only (as the sphere rolls without slipping) and doesn't cause any displacement, and the normal force is perpendicular to the displacement, so it doesn't do work.

So, we can write (for this part of the trajectory) the following expression:

ΔK + ΔU = 0 (1)

Now, if the ball starts from rest, we can say that Ki =0.

So, ΔK = Kf = 1/2 mv² + 1/2 I ω² (2)

The moment of inertia I, for  a solid sphere, is as follows:

I = 2/5 m*r²

As the ball rolls without slipping, there is a fixed relationship between angular and linear velocity, as follows:

ω = v/r, where r= radius of the sphere.

Replacing these values in (2), we have:

ΔK = 1/2 m*v² + 1/2 ((2/5*m/r²)*(v²/r²))

Simplifying and arranging terms we have:

ΔK = 7/10*m*v² (3)

For ΔU, we can write the following equation (Taking as zero reference level the bottom of the hill):

ΔU = Uf - Ui = m*g(h/2) - m*g*h = -m*g*h/2 (4)

Replacing (3) and (4) in (1):

7/10*m*v² +( -m*g*h/2) = 0

Solving for v, we get;

v = √((5/7)*g*h)

c) If at halfway point (height h/2) the hill becomes smooth, so there is no friction anymore, this means that from this point onwards, the kinetic energy of the ball is only traslational.

We can write the same equation as above, as no non-conservative forces do work:

ΔK + ΔU = 0 (1)

ΔK is just the difference between the traslational kinetic energy at the bottom of the hill, and the one for the halfway point, as there is no change in the rotational kinetic energy in this part of the path.

⇒ ΔK = Kf-K₀ = 1/2*m*vf² - 1/2 m*v₀² (2)

The change in the gravitational potential energy, is just the same as above, i.e. , we can write the following expression:

ΔU = Uf -Ui = 0- m*g*h/2 = -m*g*h/2 (3)

Replacing (2) and (3) in (1):

1/2*m*vf² - 1/2 m*v₀² + (-m*g*h/2) = 0

Replacing the value for v₀ that we got in a) and solving for vf, we get:

vf = √((12/7*g*h)

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over