A solid sphere starts from rest and rolls without slipping a distance of 8 m down a roof that is inclined at 25˚. The roof’s edge is 5 m high from the ground. How far horizontally from the roof’s edge does the sphere hit the ground?

###### Solution.

The solid sphere rolls without slipping. It means that it has linear and rotational kinetic energy.

We use conservation of energy.

The sphere’s initial kinetic energy is Ki = 0 and its initial potential energy is

Ui = Mgh, where h = 8*sin(25˚) ≈ 3.381 m.

We are using the edge of the roof as our reference level for computing U.

The sphere’s final kinetic energy as it leaves the roof is

Kf = 0.5Mv^2 + 0.5Iω^2.

We know that v = Rω or ω = v/R then

Kf = 0.5Mv^2 + 0.5I(v/R)^2.

Also, we can use the fact that I = 0.4MR^2, thus

Kf = 0.5Mv^2 + 0.2Mv^2 = 0.7Mv^2.

Now we can write the equation for the conservation of energy

Ui = Kf

or

Mgh = 0.7Mv^2

v = √(gh/0.7) = √ (9.81*3.381/0.7)≈ 6.88 m/s

We calculated the velocity of the sphere when it leaves the roof.

Now, this becomes a projectile motion.

The sphere’s velocity components are

v_{ix} = vcos(25◦) ≈ 6.235 m/s and

v_{iy} = -vsin(25◦) ≈ -2.908 m/s.

The position-time equation of vertical motion of the sphere is given by

y_{f} = y_{i} + v_{iy}t – 0.5gt^2.

We substitute the values into the equation

0 = 5 – 2.908*t – 0.5*9.81*t^2

and solve for t

t ≈ 0.756 s.

The position-time equation of horizontal motion of the sphere is given by

x_{f} = x_{i} + v_{ix}t.

We can substitute the values into the equation and solve

x_{f} = 0 + 6.235*0.756 ≈ 4.71 m

Answer: The sphere hits the ground 4.71 m horizontally from the roof’s edge.