Pad 18. 2/3 points| Previous Answers SerPOPS 5.AE 008. My Notes Ask Your Example
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Pad 18. 2/3 points| Previous Answers SerPOPS 5.AE 008. My Notes Ask Your Example 5.8 Riding the Ferris Wheel A child of mass m rides on a Ferris wheel as shown in figure (a). The child moves in a vertical circle of radius 13.5 m at a constant speed of 2.80 mys. Top (A) Determine the force exerted by the seat on the child at the bottom of the ride. Express your answer in terms of the weight of the child, mg. Boctom (B) Determine the force exerted by the seat on the child at the top of the ride. (a) Achad ndes on Ferris wtw.(b)The forces acting on the chid at the bottom of the path () The foroes acting on the did at the top of the pam (A) Determine the force exerted by the seat on the child at the bottom of the ride. Express your answer in terms of the weight of the child, mg Conceptualize Look carefully at figure (a). Based on experiences you may have had on a Ferris wheel or driving over small hills on a roadway, you would expect to feel lighter at the top of the path. Similarly, you would expect to feel heavier at the bottom of the path. At both the bottom of the path and the top, the normal and gravitational forces on the child act in opposite directions. The vector sum of these two forces gives a force of constant magnitude that keeps the child moving in a circular path at a constant speed. To yield net force vectors with the same magnitude, the normal force at the bottom must be greater than that at the top. Categorize Because the speed of the child is constant, we can categorize this problem as one involving a particle (the child) in uniform circular motion, complicated by the gravitational force acting at all times orn the child. Analyze We draw a diagram of forces acting on the child at the bottom of the ride as shown in figure (b) The only forces acting on him are the downward gravitational force ·nB and the upward force D exerted by the seat. The net upward force on the child that provides his centripetal acceleration has a magnitude o mg Apply Newton's second law to the child in the radial direction when he is at the bottom of the ride: Solve for the force exerted by the seat on the child: nbot-mg + m_.mg( 1+ Substitute the values glven for the speed and radius: (1(2.80 m/s) (13.5 mX9.80 m/s') 1.0592 mg Hence, the magnitude of the force ne exerted by the seat on the child is greater than the weight of the child.Explanation / Answer
Given,
r = 13.5 m ; v = 2.8 m/s
when the ride is midway
F = mv^2/r
multiplying and dividing the right hand side with g
F = gmv^2/rg = (mg)v^2/rg
F = 2.8^2/13.5 x 9.81 (mg) = 0.0592 (mg)
Hence, F = 0.0592 mg
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