The Windseeker ride at Cedar Point swings people in a circle at a height of 92.0

Question

The Windseeker ride at Cedar Point swings people in a circle at a height of 92.0m. The swings move at 13.0m/s (that's 30mph!) in a circle in circle of radius of 11.5m. Calculate the centripetal acceleration a rider experiences. Divide your answer by 9.8m/s^2 to determine how many "gees" a rider experiences. A rider+chair has a mass of 100.kg. What is the magnitude of their weight? If the Wind seeker spinning at constant speed, is the rider accelerating horizontally? If the Windseeker spinning at constant speed, is the rider accelerating vertically? Each chair is suspended by a metal rod. When spinning at maximum speed, the swings out at 33.7degree from the horizontal. The vertical component of the tension must balance the weight of rider+chair. Calculate the total tension. The manufacturer has a choice between using a cheaper rod that can withstand 1100N of tension and a more expensive rod that can withstand 1900N of tension. Is it safe to go with the cheaper rod? I'm trying to loosen lug nuts on a tire, but they are very tight. I can only exert 500Nn on my wrench and that is not enough torque. To increase my torque I should get a wrench with a thicker handle longer handle shorter handle thinner handle In the previous question, if I am pushing down on the wrench the torque I exert is positive or negative? Two workmen hold a 6.0m ladder. The ladder has a mass of 15kg. A cat, with a mass of 5.0kg, is on the ladder 1.0m from the edge. The ladder is in mechanical equilibrium. Assume a pivot is at the workman on the left. Draw a free body diagram for the ladder. Calculate the force of gravity the ladder feels (its weight). Gravity always acts on the center of mass of the object, so at what radius from the person the left is the center of the ladder?

Explanation / Answer

here,

d.

when the windseeker spinning at constant speed

the centripital accelration acts on the winseeker as its direction changes

so the rider is accelrating horizontally

e.

as the position of the windseeker is vertically constant

then it is not accelrated vertically

f.

let the tension be t

equating the forces vertically

t*sin(33.7) - m*g = 0

t*sin(33.7) - 100*9.8 = 0

t = 1766.26 N

the tension in the rod is 1766.26 N

g.

no its is not safe to go with the cheaper rod because the tension in the rod is more than 1100 N

and the rod will break

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