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3 years ago

Give an example of a compound machine. Explain how at least two simple machines are part of this complex machine.

Physics

1 answer:

11Alexandr11 [23.1K]3 years ago

6 0

Answer:

Bicycle

Explanation:

A compound machine is a machine which is a combination of simple machines.

Simple machines are like the pulley, inclined plane or a screw.

Suppose a bicycle is considered, it has more than one simple machine combined together, for it to work. Wheel and axle is one of them and the beam which is pivoted at a fixed hinge is another simple machine in it.

The pedals of the bicycle function as the lever.

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In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau

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the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

Em₀ = U = m g y₁

final point. At the exit of the ramp

Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

Em₀ = Em_f

m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

v = w r

the moment of inertia of a sphere is

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we substitute

m g (y₁ - y₂) = ½ m v² + ½ ( $\frac{2}{5}$ m r²) ( $\frac{v}{r}$ )²

m g h = ½ m v² (1 + $\frac{2}{5}$ )

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

mg h = $\frac{7}{5}$ ( $\frac{1}{2}$ m v²)

v = √5/7 √2gh

This is the exit velocity of the vertical movement of the sphere

v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

starting point

Em₀ = U = mg y₁

final point

Em_f = K + U = ½ m v² + m g y₂

Em₀ = Em_f

mg y₁ = ½ m v² + m g y₂

m g (y₁ -y₂) = ½ m v²

v = √2gh

this is the speed of the box

v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

v² = v₀² - 2 g y

at the highest point v = 0

y = vo₀²/ 2g

for the sphere

y_sphere = 5/7 2gh / 2g

y_esfera = 5/7 h

for the block

y_block = 2gh / 2g

y_block = h

therefore the block reaches higher than the sphere

$\frac{y_{sphere}} {y_bolck} = 5/7$

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