Last time we developed an equation to compute V, of the in our example pulley and belt system. Today we’ll plug numbers into this equation and arrive at a numerical value for this beltbelt velocity.
The equation we’ll be working with is,
t (1)_{2}where, π represents the constant 3.1416. We learned that Pulley 2’s period of revolution, t, is related to its rotational speed, _{2}N, which represents the time it takes for it to make one revolution and is represented by this equation,_{2}
÷ t_{2 }_{ }(2) We’ll now solve for the
÷ N_{2 }_{ }(3) We were previously given that revolutions per minute, so equation (3) becomes,
÷ 300 RPM = 0.0033 minutes (4)_{}This tells us that Pulley 2 takes 0.0033 minutes to make one revolution in our pulley-belt assembly. Pulley 2’s diameter,
We’ve now determined that the of 237.99 feet per minute.velocity Next time we’ll apply this value to equation (6) and determine the belt’s tight side tension, Copyright 2017 – Philip J. O’Keefe, PE Engineering Expert Witness Blog ____________________________________ |