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goldenfox [79]
3 years ago
9

A spacecraft at rest has moment of inertia of 100 kg-m^2 about an axis of interest. If a 1 newton thruster with a 1 meter moment

arm is fired orthogonal to the axis of interest for 1 minute, how far has the vehicle rotated when the firing is complete?
Physics
1 answer:
snow_lady [41]3 years ago
3 0

Answer:

18 radians

Explanation:

The computation is shown below:

As we know that

Torque = Force × Moment arm

= 1N × 1M

= 1N-M

Torque = I\alpha

\alpha = \frac{torque}{I}\\\\= \frac{1}{100}\\\\= 0.01 rad/s^2

Now

\theta = w_ot + \frac{1}{2} \alpha t^2\\\\w_o = 0\\\\\theta = 0 \times 60 + \frac{1}{2} \times 0.01 \times 60^2\\\\= 18\ radians

Here t = 1 minutes = 60 seconds

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Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.3
MAXImum [283]

Answer:

Δθ = 15747.37 rad.

Explanation:

  • The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.
  • Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:

       \omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)

  • Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:

       \omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)

  • Solving for Δθ in (2):

       \theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)

  • The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:

       \theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)

  • Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:

       \omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)

  • Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:

      \theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)

  • The total angular displacement is just the sum of (3), (4) and (6):
  • Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad
  • ⇒ Δθ = 15747.37 rad.
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Answer:

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Explanation:

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A mass m = 3.9 kg hangs from a massless string wrapped around a uniform cylinder with mass Mp = 10.53 and radius R= 0.96 m. The
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Answer:

the rotational inertia of the cylinder = 4.85 kgm²

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Explanation:

Formula for calculating Inertia can be expressed as:

I =\frac{1}{2}mR^2

For calculating the rotational inertia of the cylinder ; we have;

I = \frac{1}{2}m_pR^2

I = \frac{1}{2}*10.53*(0.96)^2

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T = \frac{Ia}{R^2}

a = \frac{g}{1+\frac{I}{mR^2}}

a = \frac{9.8}{1+\frac{4.85}{3.9*(0.96)^2}}

a = 4.1713 m/s²

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v^2 = u^2+2as \\ \\ v^2 = 2as \\ \\ v = \sqrt{2*a*s} \\ \\ v= \sqrt{2*4.1713*7.56} \\ \\ v = 7.942 \ m/s

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