Question

(B) Determine the force exerted by the seat on the child at the top of the ride. Analyze The diagram of forces acting on the child at the top of the ride is shown in figure (c). The net downward force that provides the centripetal acceleration has a magnitude mg - ntop. Apply Newton's second law to the child at this position: ?F = mg - ntop = m v²/r ntop = mg - m v²/r = mg(1 - v²/rg) ntop = mg(1 - (2.90 m/s)² / ((13.3 m)(9.80 m/s²))) ntop = mg In this case, the magnitude of the force exerted by the seat on the child is less than his true weight, and the child feels lighter. Finalize The variations in the normal force are consistent with our prediction in the Conceptualize step of the problem. Calculate the magnitude of the force on the child causing the centripetal acceleration if the ride is midway up the circle (on his way up). Express your answer in terms of mg. F = mg

          (B) Determine the force exerted by the seat on the child at the top of the ride.
Analyze The diagram of forces acting on the child at the top of the ride is shown in figure (c). The net downward force that provides the centripetal acceleration has a magnitude mg - ntop.
Apply Newton's second law to the child at this position:
?F = mg - ntop = m v²/r
ntop = mg - m v²/r = mg(1 - v²/rg)
ntop = mg(1 - (2.90 m/s)² / ((13.3 m)(9.80 m/s²)))
ntop = mg
In this case, the magnitude of the force exerted by the seat on the child is less than his true weight, and the child feels lighter.
Finalize The variations in the normal force are consistent with our prediction in the Conceptualize step of the problem.
Calculate the magnitude of the force on the child causing the centripetal acceleration if the ride is midway up the circle (on his way up). Express your answer in terms of mg.
F = mg

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(B) Determine the force exerted by the seat on the child at the top of the ride.
Analyze The diagram of forces acting on the child at the top of the ride is shown in figure (c). The net downward force that provides the centripetal acceleration has a magnitude mg - ntop.
Apply Newton's second law to the child at this position:
?F = mg - ntop = m v²/r
ntop = mg - m v²/r = mg(1 - v²/rg)
ntop = mg(1 - (2.90 m/s)² / ((13.3 m)(9.80 m/s²)))
ntop = mg
In this case, the magnitude of the force exerted by the seat on the child is less than his true weight, and the child feels lighter.
Finalize The variations in the normal force are consistent with our prediction in the Conceptualize step of the problem.
Calculate the magnitude of the force on the child causing the centripetal acceleration if the ride is midway up the circle (on his way up). Express your answer in terms of mg.
F = mg

Added by Diana A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Ryan Mcnamara

02/12/2022

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Step 1: Start with the equation for the net force in circular motion: \( F_{net} = m \frac{v^2}{r} \). Show more…

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(B) Determine the force exerted by the seat on the child at the top of the ride. Analyze The diagram of forces acting on the child at the top of the ride is shown in figure (c). The net downward force that provides the centripetal acceleration has a magnitude mg - ntop. Apply Newton's second law to the child at this position: ΣF = mg - ntop = m v²/r ntop = mg - m v²/r = mg(1 - v²/rg) ntop = mg(1 - (2.90 m/s)² / ((13.3 m)(9.80 m/s²))) ntop = mg In this case, the magnitude of the force exerted by the seat on the child is less than his true weight, and the child feels lighter. Finalize The variations in the normal force are consistent with our prediction in the Conceptualize step of the problem. Calculate the magnitude of the force on the child causing the centripetal acceleration if the ride is midway up the circle (on his way up). Express your answer in terms of mg. F = mg

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Transcript

00:02 Okay, so in this problem, they actually give you quite a bit to get started here, and they just want you to crunch the numbers, but let's explain what's happening anyways.

00:11 The net force is going to be equal to, well, what are the forces acting on in this situation? it looks like they don't show a picture here, but you've got the force of gravity acting down like this, and you've got a normal force acting up like this.

00:29 And so the net force is toward the center of the circle.

00:33 That's going to be your positive direction.

00:37 Away from the center is going to be your negative direction.

00:40 So let me write net force.

00:41 The net force is going to be the force of gravity minus force normal.

00:48 The way that they wrote it was mg minus normal at the top.

00:53 It's a different notation, but same exact thing.

00:57 In circular motion, we know that the net force is equal to mb squared over r.

01:04 Okay, so that's where this comes from.

01:07 They also, it looks like, added net top over, i'm sorry, not net top, normal at the top, over, and subtracted mv squared over r on both sides.

01:24 Okay, so we are going to be getting down to something like this, mg minus mv squared over r is equal to normal at the top.

01:35 I'm just solving for that unknown normal for.

01:39 And then lastly, factored out an m and a g.

01:48 So a little odd that they factored up the g too, but i guess that's not really a big deal.

01:54 So mg, you have a 1 from this term when you factor out mg, and then here you'd have a minus v squared over r.

02:05 But they also factored out a g that wasn't there, so that's going to go on the bottom like that...

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