The wheel consists of a thin hoop weighing 2 kg and three

The wheel consists of a thin hoop weighing 2 kg and three spokes 20 cm long and weighing 0.5 kg each. A force of 5 N is applied to the wheel rim, directed tangentially to it. It is necessary to find the moment of inertia of the hoop, the moment of inertia of the entire wheel, its angular acceleration and kinetic energy 2 s after the start of rotation.

First, let's calculate the moment of inertia of the hoop. It is determined by the formula:

$I_{\text{arr}} = \frac{mR^2}{2}$,

where $m$ is the mass of the hoop, $R$ is the radius of the hoop.

Substituting the known values, we get:

$I_{\text{обр}} = \frac{2 \cdot 0,2^2}{2} = 0,04\text{ кг}\cdot\text{м}^2$.

To calculate the moment of inertia of the entire wheel, you need to take into account the moments of inertia of the hoop and three spokes. The moment of inertia of each spoke can be calculated using the formula:

$I_{\text{spokes}} = \frac{mL^2}{12}$,

where $L$ is the length of the spoke.

Substituting the known values, we get:

$I_{\text{knitting needles}} = \frac{0.5 \cdot 0.2^2}{12} = 0.0017\text{ kg}\cdot\text{m}^2$.

Since the wheel has three spokes, the moment of inertia of all spokes is equal to:

$I_{\text{all knitting needles}} = 3 \cdot I_{\text{knitting needles}} = 0.0051\text{ kg}\cdot\text{m}^2$.

Then the moment of inertia of the entire wheel is equal to:

$I_{\text{all wheels}} = I_{\text{arr}} + I_{\text{all spokes}} = 0.04\text{ kg}\cdot\text{m}^2 + 0, 0051\text{ kg}\cdot\text{m}^2 = 0.0451\text{ kg}\cdot\text{m}^2$.

The angular acceleration of the wheel can be found using the formula:

$\tau = I \alpha$,

where $I$ is the moment of inertia, $\tau$ is the moment of force, $\alpha$ is the angular acceleration.

The moment of force acting on the wheel is equal to the force multiplied by the radius of the wheel:

$\tau = FR$.

Substituting the known values ​​and solving the equation for $\alpha$, we get:

$\alpha = \frac{\tau}{I} = \frac{FR}{I} = \frac{5\text{ Н} \cdot 0,2\text{ м}}{0,0451\text{ кг}\cdot\text{м}^2} \approx 22,2\text{ рад/с}^2$.

The kinetic energy of a rotating wheel can be calculated by the formula:

$K = \frac{1}{2}I\omega^2$,

where $\omega$ is the angular velocity of the wheel.

In 2 seconds, angular acceleration will lead to angular velocity:

$\omega = \alpha t = 22.2\text{ rad/s}^2 \cdot 2\text{ s} = 44.4\text{ rad/s}$.

Then the kinetic energy of the wheel 2 s after the start of rotation will be:

$K = \frac{1}{2}I\omega^2 = \frac{1}{2} \cdot 0,0451\text{ кг}\cdot\text{м}^2 \cdot (44,4\text{ рад/с})^2 \approx 43,7\text{ Дж}$.

Thus, we found the moment of inertia of the hoop, the moment of inertia of the entire wheel, its angular acceleration and kinetic energy 2 s after the start of rotation, when a force of 5 N is applied, directed tangentially to the wheel rim.

Welcome to our digital goods store! We are pleased to present you our new product - an exciting physics problem.

In this problem you have to calculate the moment of inertia and angular acceleration of a wheel, which consists of a thin hoop weighing 2 kg and three spokes 20 cm long and weighing 0.5 kg each. In this case, a force of 5 N is applied to the wheel rim, directed tangentially to it.

You will be able to test your knowledge of physics, apply formulas to calculate the moment of inertia and angular acceleration, and also calculate the kinetic energy of the wheel 2 seconds after the start of rotation.

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Don't miss the opportunity to purchase our digital product and test your physics knowledge!

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Given: hoop mass m₁ = 2 kg; spoke mass m₂ = 0.5 kg; knitting needle length l = 20 cm = 0.2 m; force applied to the wheel rim F = 5 N; time t = 2 s.

Let's find the moment of inertia of the hoop: I₁ = (m₁r²)/2, where r is the radius of the hoop.

Since the wheel is thin, its radius can be found from the length of the spokes: 2πr = 3l, whence r = 3l/(2π) = 0.03 m.

Then the moment of inertia of the hoop will be: I₁ = (m₁r²)/2 = (2 * 0.03²) / 2 = 0.0009 kg m².

Let's find the moment of inertia of the entire wheel: I = I₁ + ΣI₂, where ΣI₂ is the moment of inertia of all three spokes.

The moment of inertia of the spokes can be found using the formula: I₂ = (m₂l²)/12 + (m₂r²)/4, where the first term is the moment of inertia of the spokes relative to their centers of mass, and the second is the moment of inertia of the spokes relative to the axis of rotation (the center of the hoop).

The mass of one knitting needle is half the mass of the hoop, so m₂ = 0.5 kg.

Then the moment of inertia of each spoke will be: I₂ = (0.5 * 0.2²)/12 + (0.5 * 0.03²)/4 = 0.000025 kg m².

And the moment of inertia of the entire wheel: I = I₁ + ΣI₂ = 0.0009 + 3 * 0.000025 = 0.000975 kg m².

Let's find the angular acceleration of the wheel: τ = Fr, where τ is the moment of force, r is the radius of the wheel.

Since the force is applied to the rim, then r = 0.03 m.

Then the moment of force will be: τ = Fr = 5 * 0.03 = 0.15 N m.

The angular acceleration of the wheel will be: α = τ/I = 0.15/0.000975 = 153.85 rad/s².

Let's find the kinetic energy of the wheel 2 s after the start of rotation: E = (Iω²)/2, where ω is the angular speed of the wheel.

The angular velocity of the wheel 2 s after the start of rotation will be: ω = αt = 153.85 * 2 = 307.7 rad/s.

Then the kinetic energy of the wheel will be: E = (Iω²)/2 = (0.000975 * 307.7²) / 2 = 45.36 J.

Answer: moment of inertia of the hoop I₁ = 0.0009 kg m²; moment of inertia of the entire wheel I = 0.000975 kg m²; angular acceleration of the wheel α = 153.85 rad/s²; kinetic energy of the wheel 2 s after the start of rotation E = 45.36 J.

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