can someone help me with this problem than you So the summer is approaching on u

Question

can someone help me with this problem than you

So the summer is approaching on us pretty fast and with that being said, so are your house chores at home. With the Texas heat coming on, you know you're going to get stuck having to mow the lawn and rack leaves and such. Mowing the lawn doesn't exactly sound like much fun and is nowhere near the plans you have for the summer so you decide to modify your dad's lawn mower. You figure that if you double the blade system on the lawn mower, you should be able to finish a lot faster. Instead of using just the one blade on the mower, you attach a second blade perpendicular to the first. The only problem you see is if the horsepower your dads lawn mowers engine has is sufficient for the double blade system you want to install. The mowers horsepower is 6.75 HP, the length of the blade is I = 22 inches by the width w = 2.25 inches. The mass m of the blade is 11.3 kg and runs at an angular velocity of omega omega = 100 rad/s. What is the moment of Inertia of the 2 blades (treated as a rectangular plate)? How much Torque on the blades would be required by the lawn mower engine to keep the same angular velocity? Does the lawn mower provide sufficient Torque/H.P. to the blades? Explain your answer by-providing the difference in Torque/HP. (not just a yes or no answer)

Explanation / Answer

(a)

length L = 22 inch

1 inch = 2.54 cm

= 0.0254 m

L = (22) (0.0254)

= 0.55 m

width W = 2.25 inch

= (2.25) (0.0254)

= 0.05 m

the moment of inertia of each blade is m L2/ 3

so the totyal moment of inertia is

I = 2 (m L2 / 3)

= 2 (11.3) (0.55)2 / 3

= 2.27 kg m2

(b)

power = (torque) (angular velocity)

given power P = 6.75 Hp

1 Hp = 736 Watt

P = (6.75) (736)

= 4968 Watt

torqe = power / angular velocity

= (4968) / (1000)

= 4.968 rad / s2

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