Question 1199528
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A steel bar with a radius of 6 inches is being ground in a lathe. 
The surface speed of the bar is 29 ft/min. How many revolutions will the bar make in 51 sec?
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Let n be the number of revolutions per minute.

The diameter of the bar is d = 2*6 inches = 12 inches = 1 ft.


The surface speed of 29 ft/min is  the circumference  {{{pi*d}}} times the number 
of revolution per minute, so we have this equation

    29 = {{{3.14*1*n}}}.


From this equation 

    n = {{{29/3.14159}}} = 9.231  revolutions per minute.


The number of revolution in 51 seconds is  {{{(51/60)*9.231}}} = 7.846   (rounded).    <U>ANSWER</U>
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