physics, rolling + energy?

Question

a solid cylinder of radius 7 cm and mass 16 kg starts from rest and rolls without slipping a distance L = 6.0 m down a roof that is inclined at angle = 30°.

Scenario: The cylinder slides down a roof for a distance L, the roof is inclined at a 30 degree angle with the horizontal, and the roof is a height H above the ground.

(a) What is the angular speed of the cylinder about its center as it leaves the roof?
rad/s
(b) The roof's edge is at height H = 5.0 m. How far horizontally from the roof's edge does the cylinder hit the level ground?
m

Answer 1

The problem statement has a problem. If the cylinder slides down it does not roll down. If id slides down one can treat it as a body of mass (m) however when id rolls down we have to deal with a rolling mass and consider moment of inertia (I)
Where

I=0.5 m R^2

also kinetic energy of the cylinder is
Ke=0.5 I w^2
w - is angular velocity.
And since energy is energy the energy at the top of the ramp (Pe) equal to kinetic energy (Ke) at the bottom.
Ke=Pe=mgh
(a)
.5 I w^2=mgh
w=sqrt(2mgh/I)
w=sqrt(2mgh/.5 m R^2)
w=sqrt(4gh/ R^2) h=Lsin(30)=3m
w=sqrt(4 x 9.81 x 3/(.07)^2)
w=155 rad/sec
(b) In this case the angular speed has to be decomposed into vertical and horizontal components. The horizontal component will be responsible for motion away from the roof (and therefore distance) while the time in the air will be determined by the cylinders vertical movement.

I’m assuming that the cylinder had no normal (to the cylinder) linear velocity and that only tangential component was present.
In general angular speed (w) can be expressed
w=V sin (angle between normal and tangential component) /R
V- Linear component of the velocity (the outside surface of the cylinder)
R- Radius of the cylinder

Since the angle is assumed to be 90 degrees (no normal component)
w=V/R
V=w R
Let’s compute vertical and horizontal components
Vh=Vcos(30)=w R cos (30)
Vv=Vsin(30) = 0.5 w R
Time it takes fro the cylinder to hit the ground at freefall + initial velocity Vv

H=Vvt +.5 gt^2 we have
5=(0.5 w R)t + .5 gt^2
Solve for t

Then use that value (positive value) to compute the distance in question

S=Vh t

Let me know if you have any further questions.