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Alisiya [41]
3 years ago
6

A bowling ball is far from uniform. Lightweight bowling balls are made of a relatively low-density core surrounded by a thin she

ll with much higher density. A 7.0 lb (3.2 kg) bowling ball has a diameter of 0.216 m; 0.196 m of this is a 1.6 kgcore, surrounded by a 1.6 kg shell. This composition gives the ball a higher moment of inertia than it would have if it were made of a uniform material. Given the importance of the angular motion of the ball as it moves down the alley, this has real consequences for the game.
(a)Model a real bowling ball as a 0.196-m-diameter core with mass 1.6 kg plus a thin 1.6 kg shell with diameter 0.206 m (the average of the inner and outer diameters). What is the total moment of inertia?

Express your answer with the appropriate units.

(b)Find the moment of inertia of a uniform 3.2 kg ball with diameter 0.216 m.

Express your answer with the appropriate units.
Physics
1 answer:
tester [92]3 years ago
7 0

Answer:

a)  I = 1,75 10-² kg m²  and b)  I = 1.49 10⁻² kg m²

Explanation:

The expression for the moment of inertia is

    I = ∫ r² dm

The moment of inertia is a scalar by which an additive magnitude, we can add the moments of inertia of each part of the system, taking into account the axis of rotation.

    I = I core + I shell

The moment of inertia of a solid sphere is

    I sphere = 2/5 MR²

The moment of inertia of a thin spherical shell is

    I shell = 2/3 M R²

a) Let's apply to our system, first to the core of weight 1.6 kg and diameter 0.196m, the radius is half the diameter

     R = d / 2

     R= 0.196 m / 2 = 0.098 m

     I core = 2/5 1.6 0.098²

     I core = 6.147 10-3 kg m²

Let's calculate the moment of inertia of the shell of mass 1.6 kg with a diameter of 0.206 m

    R = 0.206 / 2

    R = 0.103 m

    I shell = 2/3 1.6 0.103²

    I shell = 1,132 10-2 kg m²

The moment of inertia of the ball is the sum of these moments of inertia,

    I = I core + I shell

    I = 6,147 10⁻³ + 1,132 10⁻² = 6,147 10⁻³ + 11.32 10⁻³

    I = 17.47 10⁻³ kg m²

    I = 1,747 10-² kg m²

b) Now the ball is report with mass 3.2kg and diameter 0.216 m

    R = 0.216 / 2

    R = 0.108 m

It is a uniform sphere

    I = 2/5 M R²

    I = 2/5 3.2 0.108²

    I = 1.49 10⁻² kg m²

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Part 1 :
tatuchka [14]

1) 9.57 N

We have two forces applied on the apple:

- The force of gravity, in the downward direction:

W = 9.42 N

- The force exerted by the wind, in the horizontal direction (to the right):

Fw = 1.68 N

The two forces are perpendicular to each other, so we can find the magnitude of the net force by using Pythagorean's theorem.

Therefore, we have:

F=\sqrt{W^2+F_w^2}=\sqrt{(9.42)^2+(1.68)^2}=9.57 N

2) 10^{\circ}

The direction of the net external force, measured from the downward vertical, can be measured using the following formula:

\theta = tan^{-1}(\frac{F_x}{F_y})

where

F_x is the force in the horizontal direction

F_y is the force in the vertical direction

In this problem,

F_x = F_w = 1.68 N

F_y = W = 9.42 N

and so we find:

\theta = tan^{-1}(\frac{1.68}{9.42})=10^{\circ}

4 0
2 years ago
You drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the dr
bulgar [2K]

Answer:

 y = y₀ (1 - ½ g y₀ / v²)

Explanation:

This is a free fall problem. Let's start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i

          y = y₀ + v₀ t - ½ g t²

          y = y₀ - ½ g t²

for the ball thrown from the ground with initial velocity v₀₂ = v

         y₂ = y₀₂ + v₀₂ t - ½ g t²

     

in this case y₀ = 0

         y₂2 = v t - ½ g t²

at the point where the two balls meet, they have the same height

         y = y₂

         y₀ - ½ g t² = vt - ½ g t²

         y₀i = v t

         t = y₀ / v

since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height

         y = y₀ - ½ g t²

         y = y₀ - ½ g (y₀ / v)²

         y = y₀ - ½ g y₀² / v²

        y = y₀ (1 - ½ g y₀ / v²)

with this expression we can find the meeting point of the two balls

6 0
3 years ago
Laboratory experiments, observational field studies, and model-building are all examples of different forms of scientific invest
Ne4ueva [31]

Answer:

C

Explanation:

8 0
3 years ago
A box at rest has the shape of a cube 3.5 m on a side. This box is loaded onto the flat floor of a spaceship and the spaceship t
guapka [62]

Answer:

V_o=25.725\ m^3

Explanation:

Given:

sides of the cube, a=3.5\ m

speed of the cube with respect to the observer, v=0.8c

Since the relative velocity of the object is relativistic, so there will be a length contraction according to the observer:

a_o=a\div\frac{1}{\sqrt{1-\frac{v^2}{c^2} } }

where:

a_o= observed length of the side along the direction of velocity

a_o=3.5\div\frac{1}{\sqrt{1-\frac{(0.8c)^2}{c^2} } }

a_o=2.1\ m is the observed length of the cube edge only in the direction of the velocity due to relativistic effect of length contraction.

So the observed volume will be:

V_o=a\times a\times a_o

V_o=3.5\times 3.5\times 2.1

V_o=25.725\ m^3

5 0
3 years ago
the period of a mechanical wave is 12 seconds what is the frequency of the wave 12 waves/seconds 4.0 waves/seconds 0.083 waves/s
Mandarinka [93]

Use your units ! 
"Period" is time . . . 12 seconds.
"Frequency" how often . . . " ___ PER time".  Time is in the denominator.

"12 seconds per wave" means 

                                    1 wave / 12 seconds

                             =      (  1 / 12 )  (wave/second)

                             =            0.08333... wave per second

                             =            1/12  Hz.

3 0
3 years ago
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