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?
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
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
vix = vcos(25◦) ≈ 6.235 m/s and
viy = -vsin(25◦) ≈ -2.908 m/s.
The position-time equation of vertical motion of the sphere is given by
yf = yi + viyt – 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
xf = xi + vixt.
We can substitute the values into the equation and solve
xf = 0 + 6.235*0.756 ≈ 4.71 m
Answer: The sphere hits the ground 4.71 m horizontally from the roof’s edge.