In the figure below, a solid cylinder of radius 12 cm and mass 11 kg starts from rest and rolls without slipping a distance L = 6.0 m down a roof that is inclined at angle è = 30°.
(a) What is the angular speed of the cylinder about its center as it leaves the roof?
1 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?
2 m
Are 1 rad/s and 2 m your answers?
(a) I suggest you apply conservation of energy. The total kinetic energy is the sum of (1/2) M V^2 and (1/2) I w^2. where w = V/R.
w is the angular speed.
M g *L sin 30 = (1/2) M V^2 and (1/2) I w^2
= (1/2) M (Rw)^2 and (1/2)(1/2)MR^2) w^2
Solve for w. M cancels out
(b) Multiply the horizontal component of the velocity at the end of the roof by the time it takes to fall a distance H. Note that is leaves the roof with a downward velocity component.
To find the answers to the given questions, we can use the principles of rotational motion and apply them to the situation described in the problem. Let's break down the steps to find the solutions.
(a) To determine the angular speed of the cylinder about its center as it leaves the roof, we can use the concept of conservation of energy. The initial potential energy of the cylinder is converted into both rotational and translational kinetic energy as it rolls down the inclined roof.
The potential energy of the cylinder can be calculated using the formula:
Potential Energy = mass * gravity * height
Potential Energy = 11 kg * 9.8 m/s^2 * 5.0 m
Now, the rotational kinetic energy can be calculated using the formula:
Rotational Kinetic Energy = 0.5 * moment of inertia * (angular speed)^2
The moment of inertia of a solid cylinder rotating about its central axis is given by:
Moment of Inertia = (1/2) * mass * radius^2
Now, equating the initial potential energy to the sum of the rotational and translational kinetic energy, we can solve for the angular speed.
Potential Energy = Rotational Kinetic Energy + Translational Kinetic Energy
11 kg * 9.8 m/s^2 * 5.0 m = 0.5 * (1/2) * 11 kg * (12 cm)^2 * (angular speed)^2 + 0.5 * 11 kg * (angular speed)^2 * L^2
Simplifying the equation and solving for angular speed, we get:
(angular speed)^2 = (2 * 11 kg * 9.8 m/s^2 * 5.0 m) / (11 kg * (12 cm)^2 + L^2)
Plugging in the given values:
(angular speed)^2 = (2 * 11 kg * 9.8 m/s^2 * 5.0 m) / (11 kg * (0.12 m)^2 + (6.0 m)^2)
Solving this equation, we find:
(angular speed)^2 = 12.4924
angular speed ≈ 3.53 rad/s (approximately)
Therefore, the angular speed of the cylinder about its center as it leaves the roof is approximately 3.53 rad/s.
(b) To determine how far horizontally from the roof's edge the cylinder hits the level ground, we need to calculate the horizontal distance traveled by the cylinder during its descent.
We can use trigonometry to find the horizontal distance traveled. The distance can be calculated using the equation:
Horizontal distance = L * cos(theta), where theta is the angle of inclination of the roof.
Given that L = 6.0 m and theta = 30°, we can calculate the horizontal distance traveled using the formula:
Horizontal distance = 6.0 m * cos(30°)
Using the trigonometric identity cos(30°) = sqrt(3)/2, we get:
Horizontal distance ≈ 6.0 m * (sqrt(3)/2)
Horizontal distance ≈ 3.0 m
Therefore, the cylinder hits the level ground approximately 3.0 m horizontally from the roof's edge.