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hodyreva [135]
3 years ago
14

The tensions in the rope are 12 n and 10 n . What is the angular acceleration of the pulley?

Physics
1 answer:
Alinara [238K]3 years ago
3 0

Answer:

\mathbf{\alpha = 16.06 \ rad/s^2}

Explanation:A  \ frictionless \  pulley,  \ whic h\  can \  be  \ modeled  \ as  \ a  \ 0.83 k  g \  solid \ cylinder \  with \  a \  0.30 \  m \  radius,  \  has \  a  \ rope \  going \  over  \  it The \  tensions  \ in   \ the \  rope  \ are \ 12 n \  and \  10 n .   \ What \  is \  the \  angular  \ acceleration  \ of \  the \  pulley?

From \ the \ given \ information:

The \  moment \  of  \ inerti a \  of \  a \  solid \  cylinder I = \dfrac{1}{2}mr^2

I = \dfrac{1}{2}(0.83)(0.3)^2

I = 0.03735 \ kg \ m^2

The \ net \ torque = F_1(r) - F_2(r)  \\ \\  =  12(0.3) - 10(0.3)  \\  \\ = 0.6 \ N.m

Torque (T) = moment \ of \inertia (I) \times angular \ acceleration ( \alpha)

\alpha = \dfrac{T}{I}

\alpha = \dfrac{0.6  \ Nm}{0.03735 \ kg \ m^2}

\mathbf{\alpha = 16.06 \ rad/s^2}

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Answer:

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Explanation:

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b) energy of the original photon

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