A “Gravitron” is an amusement park ride in which riders stand against the wall of a cylindrical room. When the room spins around at a sufficient rate, the floor drops down but the riders remain against the wall without sliding down. a) Thinking carefully about all the forces on a rider, draw a free body diagram for a sample rider. Draw it as a side view. Hint: there are only 3 forces. b) Write down Newton’s Second Law for both the x and y directions. The speed the ride must spin is dependent on the coefficient of static friction µ, (static instead of kinetic, because if everything goes according to plan they won’t be sliding down the wall!) and on the radius of the room r. But it can’t depend on the mass m of a rider – the speed of the ride isn’t changed based on the weights of the riders, after all. c) Solve the two equations you set up in part (b) for the speed v of the ride. Notice that the mass does actually cancel out, as promised. d) Plug in some realistic numbers: r = 2.0m and µs = 0.5. Convert the speed v you find into an “rpm” rating, revolutions per minute. Hint: one revolution is once around the circle. Treat it like a unit conversion from m/s to rev/min.
The Correct Answer and Explanation is :
Let’s go step by step for each part of this problem.
Part a) Free Body Diagram
A free body diagram is a sketch that shows all the forces acting on a rider in the Gravitron ride. Here’s a side view with the following forces:
- Gravity (Weight): The force of gravity, Fg=m⋅gF_g = m \cdot g, acts downward (toward the center of the Earth).
- Normal Force (Wall Force): The normal force, FNF_N, acts perpendicular to the wall of the ride, pushing outward from the wall.
- Static Friction: The frictional force, FfF_f, opposes the tendency of the rider to slide down the wall, and it acts upward (in the direction opposite to gravity).
The key here is that friction prevents the rider from sliding down, so the normal force and friction balance the vertical and horizontal forces.
Part b) Newton’s Second Law in x and y directions
In the Gravitron, the ride is spinning in a circular motion, so we need to apply Newton’s second law to both directions.
- In the horizontal (x) direction, we have the centripetal force that keeps the rider moving in a circle. This force comes from the normal force FNF_N because it pushes the rider toward the center of the ride. The equation is:
FN=m⋅v2rF_N = \frac{m \cdot v^2}{r}
where:
- mm is the mass of the rider,
- vv is the speed of the rider,
- rr is the radius of the circular path.
- In the vertical (y) direction, we have two forces: gravity, pulling the rider down, and static friction, which opposes gravity and prevents the rider from sliding down. At equilibrium, the static friction force equals the gravitational force:
Ff=Fg=m⋅gF_f = F_g = m \cdot g
Since the static friction is proportional to the normal force Ff=μs⋅FNF_f = \mu_s \cdot F_N, we can equate: μs⋅FN=m⋅g\mu_s \cdot F_N = m \cdot g
Part c) Solving for the speed vv
From the vertical equation, we have: μs⋅FN=m⋅g\mu_s \cdot F_N = m \cdot g
Substitute the expression for FNF_N from the horizontal equation: μs⋅m⋅v2r=m⋅g\mu_s \cdot \frac{m \cdot v^2}{r} = m \cdot g
Canceling mm from both sides (since it does not affect the result): μs⋅v2r=g\mu_s \cdot \frac{v^2}{r} = g
Solve for vv: v2=g⋅rμsv^2 = \frac{g \cdot r}{\mu_s} v=g⋅rμsv = \sqrt{\frac{g \cdot r}{\mu_s}}
Part d) Plugging in values and converting to rpm
Now let’s plug in the given values:
- r=2.0 mr = 2.0 \, \text{m}
- μs=0.5\mu_s = 0.5
- g=9.8 m/s2g = 9.8 \, \text{m/s}^2
Substitute into the equation for vv: v=9.8⋅2.00.5=19.60.5=39.2≈6.26 m/sv = \sqrt{\frac{9.8 \cdot 2.0}{0.5}} = \sqrt{\frac{19.6}{0.5}} = \sqrt{39.2} \approx 6.26 \, \text{m/s}
Now, to convert this speed into revolutions per minute (rpm), we use the fact that the speed vv is related to the circumference of the circle C=2πrC = 2 \pi r, so the number of revolutions per second is: rev/s=vC=v2πr\text{rev/s} = \frac{v}{C} = \frac{v}{2 \pi r}
Substitute the values: rev/s=6.262π⋅2.0=6.2612.566≈0.498 rev/s\text{rev/s} = \frac{6.26}{2 \pi \cdot 2.0} = \frac{6.26}{12.566} \approx 0.498 \, \text{rev/s}
To convert this to rpm: rpm=0.498⋅60≈29.88 rpm\text{rpm} = 0.498 \cdot 60 \approx 29.88 \, \text{rpm}
So the ride needs to spin at approximately 30 rpm.
Explanation (300 words)
The Gravitron ride relies on centrifugal force to keep riders against the wall. When the room spins at a sufficient speed, a rider feels pressed outward against the wall, preventing them from falling due to gravity. The key to understanding this problem is realizing that there are three forces acting on the rider: gravity, static friction, and the normal force from the wall.
The normal force is what causes the centripetal acceleration required for circular motion. This force is provided by the spinning ride, and it’s directed toward the center of the room. As the ride spins faster, the normal force increases, and this allows the rider to stay pressed against the wall. Static friction works against gravity, preventing the rider from sliding down, as friction is proportional to the normal force.
In solving for the speed, we use Newton’s second law in both the horizontal and vertical directions. By equating the forces, we see that the speed of the ride depends on the radius of the room, the coefficient of static friction, and gravity, but it does not depend on the rider’s mass, as expected. This is important because the ride’s speed is designed to work for riders of any mass.
Finally, we plugged in the numbers for the radius and friction to find that the ride must spin at about 30 rpm for riders to stay safely pressed against the wall. This speed ensures that the static friction force is strong enough to counteract gravity and prevent the rider from falling.