Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of gravity.

- Assuming negligible friction, find the speed of the riders at the bottom of its arc, given the system's center of mass travels in an arc having a radius of 14.0 m and the riders are near the center of mass.
- What is the centripetal acceleration at the bottom of the arc?
- Draw a free body diagram of the forces acting on a rider at the bottom of the arc.
- Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.
- Discuss whether the answer seems reasonable.

a) $16.6 \textrm{ m/s}$

b) $19.6 \textrm{ m/s}^2$

c) $1760 \textrm{ N}$, $F_N$ is greater than $F_g$ by a factor of 3.

d) Yes, this is reasonable since it shoudl be greater than $F_g$ in order to accelerate the rider upwards, but not so much greater as to cause injury.

### Solution video

# OpenStax College Physics, Chapter 6, Problem 21 (Problems & Exercises)

### Calculator Screenshots

*r*from the pivot and so that’s the height that it has when it's horizontal. So we can substitute

*r*in place of

*h*and we'll also multiply both sides by 2 and divide both sides by

*m*, then switch the sides around and take the square root of both sides to solve for

*v*.

*v*is going to be the square root of

*2 g r*. So it's the square root of two times 9.8 meters per second squared, times 14 meters which give 16.6 meters per second. So the centripetal acceleration will be that speed squared divided by the radius of curvature of the path. So that's 16.565 meters per second squared, divided by 14 meters giving 19.6 meters per second squared. This is a free body diagram of the rider and there's the force of the ride applies upwards, we'll call that

*F n*for normal force, and then there's the force of gravity downwards and it's intentional that I've drawn the normal force longer than gravity because this person is going to accelerate upwards. They're moving upwards really quickly at this point here. I mean -- well, I guess that's a bit the wrong way to say it. They're not really moving upwards at this point but they are accelerating upwards and sometime later they will be moving upwards. So they have some centripetal acceleration there and that's due to the fact that this normal force is greater than gravity. So, the centripetal force is the normal force upwards, minus the gravity downwards, and that's going to be mass times centripetal acceleration. We can add force of gravity to both sides. So the centripetal forces is like a net force radially speaking and both of these forces are acting radially. This one is pointing towards the center and this one is pointing away from the center. So now we have normal force is mass times centripetal acceleration plus force due to gravity and substitute

*m g*in for

*f g*and then factor out the common factor

*m*between those two terms and you have the normal force is mass times bracket centripetal acceleration plus gravitational acceleration. So 60 kilograms, mass of the rider, times 19.6 meters per second squared, centripetal acceleration, plus 9.8 meters per second squared, acceleration due to gravity, which gives 1760 newtons. If we compare that to gravity, we can divide the two. So here's the expression for the force due to the ride and divide that by

*m g*and the m's cancel and so this ratio of

*F n*over

*F g*is the same as

*a c*plus

*g*over

*g*. That's 19.6 meters per second squared plus 9.8 divided by 9.8 giving 3. So the normal force is greater than gravity by a factor of three, and that's reasonable because it should be greater since they're accelerating upwards. But it shouldn't be too much greater because that would cause injury.

## Comments

This question is ridiculous. How is an introductory Physics student realistically going to be able to answer this correctly the first time? If this question involves concepts for chapter 7, it should be there instead.

This question is ridiculous. How is an introductory Physics student realistically going to be able to answer this correctly the first time? If this question involves concepts for chapter 7, it should be there instead.