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Tpy6a [65]
3 years ago
9

A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa

l to 82.0 N is applied to the rim of the wheel. The wheel has radius 0.150 m . Starting from rest, the wheel has an angular speed of 12.8 rev/s after 3.88 s. What is the moment of inertia of the wheel?
Physics
1 answer:
ivann1987 [24]3 years ago
5 0

Answer:

The moment of inertia of the wheel is 0.593 kg-m².

Explanation:

Given that,

Force = 82.0 N

Radius r = 0.150 m

Angular speed = 12.8 rev/s

Time = 3.88 s

We need to calculate the torque

Using formula of torque

\tau=F\times r

\tau=82.0\times0.150

\tau=12.3\ N-m

Now, The angular acceleration

\dfrac{d\omega}{dt}=\dfrac{12.8\times2\pi}{3.88}

\dfrac{d\omega}{dt}=20.73\ rad/s^2

We need to calculate the moment of inertia

Using relation between torque and moment of inertia

\tau=I\times\dfrac{d\omega}{dt}

I=\dfrac{I}{\dfrac{d\omega}{dt}}

I=\dfrac{12.3}{20.73}

I= 0.593\ kg-m^2

Hence, The moment of inertia of the wheel is 0.593 kg-m².

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The angular momentum of a flywheel having a rotational inertia of 0.142 kg·m2 about its central axis decreases from 7.20 to 0.14
luda_lava [24]

Answer:

3.3619 Nm

54.27472 rad

182.46618 J

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Explanation:

L_i = Initial angular momentum = 7.2 kgm²/s

L_f = Final angular momentum = 0.14 kgm²/s

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t = Time taken

Average torque is given by

\tau_{av}=\frac{L_f-L_i}{\Delta t}\\\Rightarrow \tau_{av}=\frac{0.14-7.2}{2.1}\\\Rightarrow \tau_{av}=-3.3619\ Nm

Magnitude of the average torque acting on the flywheel is 3.3619 Nm

Angular speed is given by

\omega_i=\frac{L_i}{I}

Angular acceleration is given by

\alpha=\frac{\tau}{I}

From the equation of rotational motion

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=\frac{L_i}{I}\times t+\frac{1}{2}\times \frac{\tau}{I}\times t^2\\\Rightarrow \theta=\frac{7.2}{0.142}\times 2.1+\frac{1}{2}\times \frac{-3.3619}{0.142}\times 2.1^2\\\Rightarrow \theta=54.27472\ rad

The angle the flywheel turns is 54.27472 rad

Work done is given by

W=\tau\theta\\\Rightarrow W=-3.3619\times 54.27472\\\Rightarrow W=-182.46618\ J

Work done on the wheel is 182.46618 J

Power is given by

P=\frac{W}{t}\\\Rightarrow P=\frac{-182.46618}{2.1}\\\Rightarrow P=-86.88\ W

The magnitude of the average power done on the flywheel is 86.88 W

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Answer:

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K= proportionality constant

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then we use the equation to calculate the woman's weight with the new distance

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