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bezimeni [28]
3 years ago
9

Yo-Yo man releases a yo-yo from rest and allows it to drop, as he keeps the top end of the string stationary. The mass of the yo

-yo is 0.056 kg, its moment of inertia is 2.9×10−5kg⋅m2 and the radius, r, of the axle the string wraps around is 0.0064 m.
Through what height must the yo-yo fall for its linear speed to be 0.75 m/s ?
Physics
1 answer:
patriot [66]3 years ago
6 0

To solve this problem it is necessary to use the conservation equations of both kinetic, rotational and potential energy.

By definition we know that

KE + KR = PE

Where,

KE =Kinetic Energy

KR = Rotational Kinetic Energy

PE = Potential Energy

In this way

\frac{1}{2} mv^2 +\frac{1}{2} I\omega^2 = mgh

Where,

m = mass

v= Velocity

I = Moment of Inertia

\omega = Angular velocity

g = Gravity

h = Height

We know as well that \omega = v/r for velocity (v) and Radius (r)

Therefore replacing we have

\frac{1}{2} mv^2 +\frac{1}{2} I\omega^2 = mgh

[tex]h= \frac{1}{2} \frac{v^2}{g} +\frac{1}{2} \frac{I}{mg}(\frac{v}{r})^2[/tex]

h= \frac{1}{2}v^2 ( \frac{1}{g} +\frac{I}{mg}\frac{1}{r^2} )

h= \frac{1}{2}0.75^2 ( \frac{1}{9.8} +\frac{2.9*10^{-5}}{(0.056)(9.8)}\frac{1}{(0.0064)^2} )

h = 0.3915m

Therefore the height must be 0.3915 for the yo-yo fall has a linear speed of 0.75m/s

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Answer:

increasing the separation between the plates

Explanation:

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Answer:

x = 11.23  m

Explanation:

For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.

Let's reduce to SI system units

    θ = 155 rev (2pi rad / rev) = 310π rad

    α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²

Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)

    w² = w₀² + 2 α θ  

    w =√ 2 α θ

    w = √(2 4pi 310pi)

    w = 156.45  rad / s

The relationship between angular and linear velocity

    v = w r

    v = 156.45  0.175

    v = 27.38 m / s

In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive

    y = v_{oy} t - ½ g t²

As it leaves the highest point its speed is horizontal

   y = 0 - ½ g t²

   t = √ (-2y / g)

   t = √ (-2 (-0.820) /9.8)

   t = 0.41 s

With this time we calculate the horizontal distance, because the constant horizontal speed

   x = vox t

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Margarita [4]

The change in momentum is 5500 kg m/s

Explanation:

The change in momentum of an object is given by

\Delta p = m(v-u)

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In this problem, we have:

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v=28.0 m/s (final velocity)

u=6.0 m/s (initial velocity)

Therefore, the change in momentum is

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Learn more about change in momentum:

brainly.com/question/9484203

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