Question

A hollow sphere of radius 0.15 m, with rotational inertia I 0.040 kg m2 about a line through its center of mass, rolls without slipping up a surface inclined at 30 to the horizontal. At a certain initial position, the sphere’s total kinetic energy is 20 J. (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved 1.0 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?

Answer 1

Answer:

Explanation:

Let v be the velocity of the c.m of sphere at the initial position .

Total kinetic energy = linear + rotational kinetic energy

= 1/2 mv² + 1/2 I ω² ; m is mass , I is moment of inertia and ω be angular velocity of sphere .

Total kinetic energy = 1/2 mv² + 1/2 x 1/2 m r² ω² ; (   I = 1/2 mr² , r is radius of the sphere)

Total kinetic energy = 1/2 mv² + 1/2 x 1/2 m r² ω²

= 1/2 mv² + 1/4 m v² (v = ω r , for perfect rolling )

= 3 /4 mv² = 20 J

mv² = 80 / 3

1/4 m v ² = 20 / 3

rotational kinetic energy = 20/ 3   J

b )mv² = 80 / 3

.04 v² = 80/3

v² = 666.67

v = 25.82 m /s

c )

When the sphere has moved 1.0 m up the incline from its initial position

height achieved h = 1 / sin30

= 2 m

potential energy attained by sphere = mgh

= .04 x 9.8 x 2

= .784 J

Total kinetic energy at the final position

= 20 - .784

= 19.216 J

d )

So total energy = 3 / 4 m v²

3 / 4 m v² = 19.216

3/4 x .04 v² = 19.26

v² = 642

v = 25.34 m /s

This is velocity at the final position.

Answer 2

The concepts of linear and rotational energy allow to find the results for the questions about the motion of the hollow sphere in the plane are:

     a) Rotational kinetic energy is: K = 8 J

     b) The initial velocity of the center of mass is: v = 9 m / s

     c) The final kinetic energy K = 6.93 J and the speed of the sphere at this point is v = 1.77 m / s

Given parameters

• The radius of the hollow sphere is: r = 0.15 m

• The moment of inertia I = 0.040 kg m²

• Angle of the plane tea = 30º

• The initial kinetic energy Ko = 20 J

• Displacement L = 1.0 m

To find

   a) The initial endowment kinetic energy.

   b) The initial speed of the center of mass.

   c) The kinetic energy and final velocity.

a) The kinetic energy is the energy due to the movement of the particle, it has two components linear and rotational.

           $K = K_{linear} + K_{rotational}$  

           K = ½ m v² + ½ I w²

Where K is the kinetic energy, m the mass, I the moment of inertia, the linear velocity, and the angular velocity.

 

The linear and angular variables are related.

         v = w r

         w = $\frac{v}{r}$  

The tabulated moment of inertia for a hollow sphere is:

        I = ⅔ m r²

        m = $\frac{3}{2} \ \frac{I}{r^2}$

Let's substitute.

       K = ½ ($\frac{3}{2} \ \frac{I}{r^2}$ ) (w r)² + ½ I w²

       K = ¾ I w² + ½ I w²

       K = ½ I w² ( $\frac{3}{2} + 1$ )

       K = $K_{rotational}$   $\frac{5}{2}$  

       $K_{rotational}$  = $\frac{2}{5}$ K

       

Let's calculate

       K_{rotational} = 2/5 20

       K_{rotational} = 8 J

b) let's find the linear kinetic energy.

       $K_{linear} = K - K_{rotational}$  

       $K_{linear}$  = 20 - 8

       $K_{linear}$  = 12 J

     K_{linear} = ½ m v²

      $v = \sqrt{\frac{2 K_{linear} }{m} }$

Let's calculate the mass

      m = $\frac{3}{2} \ \frac{I}{r^2}$

      m = $\frac{3}{2} \ \frac{0.040}{0.15^2}$

      m = 2,667 kg

Let's calculate the speed

       v² = $\frac{2 \ 12}{2.667}$  

       v = 9 m / s

c) Let's use the conservation of energy that states that if there is no friction, the mechanical energy is constant.

Starting point. At the point of departure.

        Em₀ = K

        Em₀ = 20 J

Final point. When L = 1.0 m has been displaced.

        $Em_f$  = K + U.

        $Em_f$ = K + m g h

Energy is conserved.

         $Em_0 = Em_f$  

         20 = K + m g h

         K = 20- m g h

Let's use trigonometry to find when the hollow sphere has risen.

         sin 30 = ha / La

         h = L sin 30

         h = 1 is 30

         h = 0.5 m

Let's calculate the kinetic energy

         K = 20 - 2.667  9.8  0.5

         K = 6.93 J

This kinetic energy has a linear and a rotational part, solving as in part a

        $K = K_{linear} + K_{rotational}$  

         K = ½ m v² + ½ (⅔ mr²) ($\frac{v}{r}$ ) ²

         K = m v² (½ + ⅓)

         K = 5/6 m v²

         v = $\sqrt{\frac{6}{5} \ \frac{K}{m} }$

let's calculate.

         v = $\sqrt{\frac{6}{5} \ \frac{6.93}{2.667} }$  

         v = 1.77 m / s

In conclusion using the concepts of linear and rotational energy we can find the results for the questions about the motion of the hollow sphere in the plane are:

     a) Rotational kinetic energy is: K = 8 J

     b) The initial velocity of the center of mass is: v = 9 m / s

     c) The final kinetic energy K = 6.93 J and the speed of the sphere at this point is v = 1.77 m / s

Learn more here:  brainly.com/question/15982753