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velikii [3]
3 years ago
14

g The tires of a car make 75 revolutions as the car reduces its speed uniformly from 91 km/h to 48 km/h . The tires have a diame

ter of 0.78 m . Part A What was the angular acceleration of the tires
Physics
1 answer:
aleksandrvk [35]3 years ago
5 0

Answer:

   α = -0.01625 rad / s²

Explanation:

This is an exercise in angular kinematics, we can use the relation

         w = w₀ + 2 α θ

linear and angular variables are related

        v = w r

         w = v / r

Let's reduce the magnitudes to the SI system

         v₀ = 91 km / h (1000m / 1km) (1h / 3600s) = 25.278 m / s

         v = 48 km / h = 13,333 m / s

         θ  = 75 rev (2π rad / 1 rev) = 471.24 rad

Let's find the angular velocities

         w₀ = v₀ / r

         w₀ = 25.278 / 0.78

         w₀ = 32,408 rad / s

         w = v / r

         w = 13.333 / 0.78

         w = 17.09 rad / s

we calculate the angular acceleration

         α = (w- w₀) / 2θ  

         α = (17.09 - 32.408) / (2 471.24)

         α = -0.01625 rad / s²

the negative sign indicates that the wheel is stopping

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A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen
PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})

Where:

\omega_{o}, \omega - Initial and final angular velocities, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

\theta_{o}, \theta - Initial and final angular position, measured in radians.

Then,

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 0\,\frac{rad}{s}, \omega = 0.70\,\frac{rad}{s} and \theta-\theta_{o} = 4.9\,rad, the angular acceleration is:

\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}

\alpha = 0.05\,\frac{rad}{s^{2}}

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

\omega = \omega_{o} + \alpha \cdot t

Where t is the time measured in seconds.

The time is cleared and obtain after replacing every value:

t = \frac{\omega-\omega_{o}}{\alpha}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and \alpha = 0.05\,\frac{rad}{s^{2}}, the required time is:

t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }

t = 14\,s

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

\bar \alpha = \frac{\omega-\omega_{o}}{t}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and t = 14\,s, the average angular acceleration is:

\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}

\bar \alpha = 0.05\,\frac{rad}{s^{2}}

The average angular acceleration is 0.05 radians per square second.

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