In the figure, a solid cylinder of radius 16 cm and mass 15 kg starts from rest

Question

In the figure, a solid cylinder of radius 16 cm and mass 15 kg starts from rest and rolls without slipping a distance L = 5.8 m down a roof that is inclined at angle theta = 21

In the figure, a solid cylinder of radius 16 cm and mass 15 kg starts from rest and rolls without slipping a distance L = 5.8 m down a roof that is inclined at angle theta = 21A degree. (a) What is the angular speed of the cylinder about its center as it leaves the roof? (b) The roof's edge is at height H = 5.2 m. How far horizontally from the roof's edge does the cylinder hit the level ground?

Explanation / Answer

Use the LCE(Law of Conservation of Energy) to determine the velocity with which the cylinder fall off the roof.

m = mass = 15kg
l = length of the roof = 5.8m
r= radius of cylinder = 16 cm =0.16 m
z= angle of roof = 21 degrees
K= kinetic energy (linear and rotational)
U= potential energy

Kb -Ka + Ub - Ua =0 (using LCE where b is bottom of the roof where a is top of the roof)
Kb =Ua=mgl sin z =305.85 J

Kb= 0.5 Iw^2 + 0.5 mv^2
I=.5*MR^2
w= v/R
Kb= 0.5[(0.5*MR^2)(v^2/R^2)] +0.5mv^2
Kb= 0.75mv^2
v= ?[4Kb/(3m)] = 1.346 m/s

A) What it the angular speed of the cylinder (I=.5*MR^2) about its center as it leaves the roof?
w= v/r =1.336/.16=8.4125 rad /s

vy= v cos z =1.256 m/s
vx = v sin z =0.482 m/s
g= gravity = 9.81 m/s^2
h = height = 5.2m
Vf= final velocity
vo= initial velocity = vy = 1.256
vf^2 = vo^2 +2gh
vf= ?(vo^2 +2gh) = 10.178 m/s
vf= vo+ gt
t=time= vf- vo /g = 0.9094 s

B) If the outside wall of the house is 5.2 m high, how far from the edge of the roof does the cylinder hit the level ground?
s=distance = vx*t = 0.438 m

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