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Ket [755]
1 year ago
4

A Ferris wheel is moving at an initial angular velocity of 1.0 rev/59 s. If the operator then brings it to a stop in 3.1 min, wh

at is the angular acceleration of the Ferris wheel
Physics
1 answer:
GalinKa [24]1 year ago
5 0

Answer:

\alpha = 5.7 * 10^{-4} rad/s^2

Explanation:

Given

w_0 = 1.0/59 \ rev/s initial angular velocity

t = 3.1\ min -- Initial time

w = 0 --- final angular velocity (when the wheel stops)

Required:

Determine the angular acceleration (\alpha)

The angular is calculated using the following formula

w = w_0 + \alpha * t

Convert time to seconds:

t = 3.1\ min

t = 3.1 * 60s

t = 186s

Convert angular velocity to rad/s

w_0 = 1.0/59\ rev/s

w_0 = \frac{1}{59} * 6.283rad/s

w_0 = \frac{6.283}{59}\ rad/s

Substitute in the required values, the expression becomes:

w = w_0 + \alpha * t

0 = \frac{6.283}{59} + \alpha * 186

0 = \frac{6.283}{59}+ 186\alpha

Collect Like Terms

186\alpha = -\frac{6.283}{59}

Make \alpha the subject

\alpha = \frac{6.283}{59 * 186}

\alpha = \frac{6.283}{10974}

\alpha = 0.00057253508rad/s^2

\alpha = 5.7 * 10^{-4} rad/s^2

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A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T
Serga [27]

Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

      k_{e} = ½ 2900 0.80²

      k_{e} = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

     U = m h and

     U = 8 9.8 (-0.80)

     U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

      K = 0

d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

     E_{mf} = mg y

     Emo = E_{mf}

     ½ k x² = m g y

     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

     Emo = ½ k x² = 928 J

     E_{mf}= ½ m v²

     Emo = E_{mf}

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s

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Explanation:

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