Question

In the figure, a solid cylinder of radius 5.3 cm and mass 17 kg starts from rest and rolls without slipping a distance L = 7.6 m down a roof that is inclined at angle θ = 22°. (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 = 3.8 m. How far horizontally from the roof's edge does the cylinder hit the level ground?

Answer 1

here,

mass , m = 17 kg

radius , r = 5.3 cm = 0.053 m

L = 7.6 m

theta = 22 degree

a)

let the angular speed at the bottom of roof be w

using conservation of energy

0.5 * m * v^2 + 0.5 * I * w^2 = m * g * (L * sin(theta))

0.5 * m * (r * w)^2 + 0.5 * (0.4 * m * r^2) * w^2 = m * g * (L * sin(theta))

0.7 * 0.053^2 * w^2 = 9.81 * 7.6 * sin(22)

solving for w

w = 119.2 rad/s

the angular speed is 119.2 rad/s

b)

the speed of ball when it leaves the roof , v = r * w

v = 0.053 * 119.2 m/s = 6.32 m/s

let the time taken to hit the ground be t

H = v * sin(theta) * t + 0.5 * g * t^2

3.8 = 6.32 * sin(22) * t + 0.5 * 9.81 * t^2

solving for t

t = 0.67 s

the horizontal distance travelled , x = v * cos(theta) * t

x = 6.32 * cos(22) * 0.67 m = 3.93 m