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coldgirl [10]
3 years ago
14

A thin spherical shell has a radius of 0.70 m. An applied torque of 860 N m gives the shell an angular acceleration of 4.70 rad/

s2 about an axis through the center of the shell. What is the rotational inertia of the shell about the axis of rotation
Physics
1 answer:
Artyom0805 [142]3 years ago
7 0

Answer:

I=182.97\ kg-m^2

Explanation:

Given that,

Radius of a spherical shell, r = 0.7 m

Torque acting on the shell, \tau=860\ N

Angular acceleration of the shell, \alpha =4.7\ m/s^2

We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

\tau=I\alpha

I is the rotational inertia of the shell

I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{860}{4.7}\\\\I=182.97\ kg-m^2

So, the rotational inertia of the shell is 182.97\ kg-m^2.

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A woman with a mass of 52.0 kg is standing on the rim of a large disk that is rotating at an angular velocity of 0.470 rev/s abo
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Explanation:

It is given that,

Mass of the woman, m₁ = 52 kg

Angular velocity, \omega=0.47\ rev/s=2.95\ rad/s

Mass of disk, m₂ = 118 kg

Radius of the disk, r = 3.9 m

The moment of inertia of woman which is standing at the rim of a large disk is :

I={m_1r^2}

I={52\times 3.9^2}

I₁ = 790.92 kg-m²

The moment of inertia of of the disk about an axis through its center is given by :

I_2=\dfrac{m_2r^2}{2}

I_2=\dfrac{118\times (3.9)^2}{2}

I₂ =897.39 kg-m²

Total moment of inertia of the system is given by :

I=I_1+I_2

I=790.92+897.39

I = 1688.31 kg-m²

The angular momentum of the system is :

L=I\times \omega

L=1688.31 \times 2.95

L=4980.5\ kg-m^2/s

So, the total angular momentum of the system is 4980.5 kg-m²/s. Hence, this is the required solution.

8 0
3 years ago
3. A coil of 100 turns encloses an area of 100 cm2. It is placed at an angle of 700 with a
sasho [114]

Explanation:

Given that,

Number of turns in the coil, N = 100

Area of the coil, A = 100 cm² = 0.01 m²

It is placed at an angle of 70°.

Magnetic field, B = 0.1 Wb/m²

We need to find the magnetic flux through the coil and the emf is induced in the coil after 10⁻³ s.

Magnetic flux is given by :

\phi =BA\cos\theta\\\\\text{For N turns},\\\phi =NBA\cos\theta \\\\\phi=100\times 0.1\times 0.01\times \cos(70)\\\\=0.034\ Wb

So, the magnetic flux through the coil is 0.1 Wb.

Emf induced in the coil is :

\epsilon=\dfrac{-d\phi}{dt}\\\\=\dfrac{0.034}{10^{-3}}\\\\=34\ V

So, 34V of emf is induced in the coil.

7 0
3 years ago
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