Question

a solid cylinder of radius 10 cm and mass 12 kg starts from rest and rolls without slipping a distance L = 6.0 m down a roof that is inclined at the angle theta = 30degree.
(a) What is the angular speed of the cylinder about its center as it leaves the roof?
(b) The roofs edge is at height H = 5.0 m. How far horizontally from the roof's edge does the cylinder hit the level ground?

Answer 1

Answer:

Explanation:

Acceleration of cylinder

a = g sin 30 / 1+ k² / r² where k is radius of gyration and r is radius of cylinder.

For cylinder k²  = (1 / 2)  r²

acceleration

= gsin30 / 1.5

= g / 3

= 3.27

v² = u² + 2as

= 2 x 3.27 x 6

v = 6.26 m /s

v = angular velocity x radius

6.26 = angular velocity x .10

angular velocity  = 62.6 rad / s

b ) vertical component of velocity

= 6.26 sin 30

= 3.13 m /s

h = ut + 1/2 g t²

5 = 3.13 t + .5 t²

.5 t²+ 3.13 t- 5 = 0

t = 1.32 s

horizontal distance covered

= 6.26 cos 30 x 1.32

= 7.15  m

Answer 2

The conservation of energy and kinematics allows to find the results for the questions about the movement of the cylinder on the ceiling and when falling are:  

          a) The angular velocity is w = 6.26 rad / s

          b) the distance to the ground is: x = 7,476 m

Given parameters

• Cylinder radius r = 10 cm = 0.10 m

• Mass m = 12 kg

• Distance L = 6.0 m

• Roof angle θ = 30º

• Ceiling height H = 5.0 m

To find

a) The angular velocity.

b) Horizontal distance.

Mechanical energy is the sum of kinetic energy and potential energies. If there is no friction, it remains constant at all points.

Linear and rotational kinematics study the motion of bodies with linear and rotational motions.

a) Let's write the mechanical energy at the points of interest.

Starting point. When it comes out of the top

          Em₀ = U = m g h

Final point. On the edge of the roof.

          $Em_f$  = K = ½ mv² + ½ I w²

Since the cylinder does not slide, friction is zero and energy is conserved.

         Em₀ = $Em_f$  

         mg h = ½ m v² + ½ I w²

The moment of inertia of the cylinder is;

        I = ½ m r²

Linear and angular variables are related.

        v = w r

let's substitute.

         m g h = ½ m (wr) ² + ½ (½ m r²) w²

        gh = ½ w² r² (1 + ½) = ½ w² r² $\frac{3}{2}$  

        w² = $\frac{4}{3 } \ \frac{gh}{r^2}$  

Let's use trigonometry to find the height of the ceiling.

        sin θ = h / L

        h = L sin θ

We substitute.

       w=    $\sqrt{ \frac{4}{3} \ \frac{g \ L sin \theta }{r^2} }$

Let's calculate.

       w = $\sqrt{\frac{4}{3} \frac{9.8 \ 6.0 sin 30}{0.10^2} }$

Let's calculate

        w = Ra 4/3 9.8 6.0 sin 30 / 0.10²

        w = 62.6 rad / s

b) For this part we can use the projectile launch expressions.

Let's find the time it takes to get to the floor.

         y = y₀ + go t - ½ g t²

The initial height is y₀ = H, when it reaches the ground its height is y = 0 and let's use trigonometry for the vertical initial velocity.

        sin  30 = $\frac{v_{oy}}{v_o}$I go / v

      $v_{oy}$  = v sin 30 = wr sin 30

       $v_{oy}$  = 62.6 0.1 sin 30

       $v_{oy}$ = 3.13 m / s

       0 = H + voy t - ½ g t²

 

       0 = 5 + 3.13 t - ½ 9.8 t³

        t² - 0.6387 t - 1.02 = 0

We solve the quadratic equation.

         t =$\frac{0.6387 \pm \sqrt{0.6387^2 - 4 \ 1.02} }{2}$  

         t = $\frac{0.6378 \pm 2.118}{2}$

         t₁ = 1.379 s

         t₂ = -0, 7 s

The time o must be a positive quantity, therefore the correct answer is:

           t = 1.379 s

We look for the horizontal distance.

          x = v₀ₓ t

          vₓ = v cos θ

          v = wr

Let's substitute.

          x = wr cos t

Let's calculate.

          x = 62.6 0.1  cos 30   1.379

          x = 7.476 m

In conclusion using the conservation of energy and kinematics we can find the results for the questions about the movement of the cylinder on the ceiling and when falling are:  

         a) The angular velocity is w = 6.26 rad / s

         b) the distance to the ground is: x = 7,476 m

Learn more here:  brainly.com/question/13949051