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juin [17]
3 years ago
6

A potter's wheel rotating at 215 rev/min is switched off and rotates through 75.0 revolutions prior to coming to rest. What is t

he constant angular acceleration of the potter's wheel during this interval
Physics
1 answer:
vaieri [72.5K]3 years ago
6 0

Explanation:

Given :

Initial angular velocity =w_i= 215 rev/min

Final angular velocity =w_f= 0

Number of revolutions=n =75.0 rev

Using relation

w_f^2-w_i^2=2\alpha n

0-215^2=2\times 75\alpha

-308.1666 rev/min^2=\alpha

Converting to rad/s^2

\alpha =-1.45 rad/s^2

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Answer:

a

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b

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Explanation:

From the question we are told that

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    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

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So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

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=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

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