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Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sphere 2.spherical shell 3.hoop 4.cylinder m/s

Respuesta :

cjrodriguez89

Answer:

1. 14.4 m/s  2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:  

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

 ½ *m*v² + ½* I* ω²  = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

  •  Solid Sphere I = 2/5* m *r²

                Replacing in (1):

                ½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(10/7*9.8 m/s2*14.7 m)  = 14. 4 m/s

  • Spherical shell I=2/3*m*r²

                Replacing in (1):

                ½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(6/5*9.8 m/s2*14.7 m)  = 13.2 m/s

  • Hoop   I= m*r²

                Replacing in (1):

                ½ * v² + ½ (m*r²) *(v/r)² = g*h

               Replacing by the value given for h, and solving for v:

               v = √(9.8 m/s2*14.7 m)  = 12.0 m/s

  • Cylinder I = 1/2 * m* r²

                 Replacing in (1):

                ½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

                 Replacing by the value given for h, and solving for v:

                v = 2*√(1/3*9.8 m/s2*14.7 m)  = 13.9 m/s

onyebuchinnaji

(1) The speed at the bottom of the solid sphere is 14.35 m/s.

(2) The speed at the bottom of the spherical shell is 13.14 m/s.

(3) The speed at the bottom of the hoop is 12 m/s.

(4) The speed at the bottom of the cylinder is 13.86 m/s.

Speed of the objects

The speed at the bottom of each object will be determied by applying the principle of conservation of energy.

K.E = P.E

¹/₂mv² + ¹/₂Iω² = mgh

¹/₂mv² + ¹/₂I(v/r)² = mgh

where;

  • I is the moment of inertia of each object

For solid sphere

I = ²/₅mr²

¹/₂mv² + ¹/₂(²/₅mr²)(v/r)² = mgh

¹/₂v² + ¹/₅v² = gh

7v² = 10gh

v² = 10gh/7

v = √10gh/7

v = √(10 x 9.8 x 14.7 / 7)

v = 14.35 m/s

For spherical shell

I = ²/₃mr²

¹/₂mv² + ¹/₂(²/₃mr²)(v/r)² = mgh

¹/₂v² + ¹/₃v² = gh

5v² = 6gh

v = √(6gh/5)

v = √(6 x 9.8 x 14.7 /5)

v = 13.14 m/s

For hoop

I = mr²

¹/₂mv² + ¹/₂(mr²)(v/r)² = mgh

¹/₂v² + ¹/₂v²  = gh

v² = gh

v = √gh

v = √(9.8 x 14.7)

v = 12 m/s

For cylinder

I = ¹/₂mr²

¹/₂mv² + ¹/₂(¹/₂mr²)(v/r)² = mgh

¹/₂v² + ¹/₄v² = gh

6v² = 8gh

v = √(8gh/6)

v = √(8 x 9.8 x 14.7 / 6)

v = 13.86 m/s

Learn more about moment of inertia of objects here: https://brainly.com/question/15248039

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