Question

A solid cylinder of radius 10 cm and mass 13 kg starts from rest and rolls without slipping a distance of 6.0 m down a house roof that is inclined at 30°. (See the figure.) What is the angular speed of the cylinder about its center as it leaves the house roof? 

The outside wall of the house is 5 m high. How far from the edge of the roof does the cylinder hit the level ground?

Answer 1

Given that

A solid cylinder of radius (R) =10 cm =0.1m

The mass of the cylinder (m) = 13 kg

The distance rolled by the solid cylinder is (s) = 6.0 m down a house roof that is inclined at (theta) =30°

The initial velocity of the solid cylinder is (u) =0

The outside wall of the house is (hi) =5 m high.

Acceleration due to gravity (g) =9.81m/s²

The acceleration of the solid cylinder is given by

a =gsintheta/1+(K/R)² =(2/3)gsinthea =(2/3)(9.81m/s2)sin30 =3.27m/s2

The speed of the cylinder when it leaves the roof is given by

v²=u²+2as

v² =2as =2*(3.27m/s2)(6.0m) ===>v =Sqrt(2*(3.27m/s²)(6.0m) =6.26m/s

a)

Now the angualr speed of the cylinder when it leaves the roof is

v =Rw

Then w =v/R =6.26m/s/0.1m =62.6rad/s

b)

The vertical height of the cylinder is (hi) =5m

Final vertical height is hf =0

The horizontal component of velocity when it leaves the roof is

vx =vocos30 =(6.26m/s)cos30 =5.42m/s

The vertical component of velocity of the ball when it leaves the roof is

vy =-vsin30 =-6.26sin30 = -3.13m/s

Now we know thata

hf-hi =vyt -(1/2)gt2

0-5m =(-3.13)t -(1/2)(9.81)t²

4.9t²+3.13t-5 =0

Now solving for t we get t =0.74s

Now the horizontal distance travelled by the cylinder is

x =vx*t =(5.42m/s)(0.74s)=4.011m