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Alex787 [66]
2 years ago
11

A bucket filled with water has a mass of 70kg and is hanging from a rope that is wound around a 0.054 m radius stationary cylind

er. If the cylinder does not rotate and the bucket hangs straight down, what is the magnitude of the torque the bucket produces around the center of the cylinder?
Physics
1 answer:
guajiro [1.7K]2 years ago
3 0

Answer:

T = 37.08 [N*m]

Explanation:

We must remember that torque is defined as the product of a force by a distance. This distance is measured from the point of application of force to the center of rotation of the rotating body.

The force is equal to the product of mass by gravitational acceleration.

F=m*g\\F=70*9.81\\F=686.7[N]

Now the torque can be calculated:

T=F*r\\T=686.7*0.054\\T=37.08[N*m]

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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
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The average angular acceleration is 0.05 radians per square second.

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\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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