Video: Determining the Centripetal Force Provided by a Normal Reaction

Ed Burdette

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00-meter radius, at how many revolutions per minute are the riders subjected to a centripetal acceleration equal to that of gravity?

06:35

Video Transcript

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00-meter radius, at how many revolutions per minute are the rider’s subjected to a centripetal acceleration equal to that of gravity?

In this problem, we’ll assume that the acceleration due to gravity is exactly 9.8 meters per second squared. In this problem statement, some of the vital information given is that the path that the riders follow on this ride has a radius of 8.00 meters.

We want to solve for the number of revolutions per minute that the riders must experience in order to be subjected to a centripetal acceleration that equals the magnitude of gravitational acceleration. Let’s refer to that number of revolutions per minute as capital 𝑁. Let’s draw a diagram of the situation.

If we look at the flying saucer-shaped ride top down from an aerial view, then we know that it looks like a disk that rotates about its center. The radius of that disc is given as 8.00 meters. And the disc rotates with some rotational speed we’ve called 𝑁. 𝑁 will be determined by the acceleration of a point on the circumference of this disc where a rider might be seated, having a centripetal acceleration that equals that of gravity.

Let’s begin by recalling the equation for centripetal acceleration. The centripetal acceleration of an object is equal to the speed of that object squared divided by the distance from that object to the center of rotation called π‘Ÿ, the radius.

So if we imagine riders seated at the edge of our rotating disc, the distance of those riders from the center of rotation is π‘Ÿ. And they have some rotational speed that we’re solving for.

In this problem, we’re told that the centripetal acceleration of the riders matches a certain constraint. We’re told that that centripetal acceleration is equal to 𝑔, the acceleration due to gravity. When we insert 𝑣 squared over π‘Ÿ for π‘Ž sub 𝑐 in our equation, we see that we know the value for π‘Ÿ, given in our problem, but 𝑣 squared is something we’ll need to determine.

Now notice that 𝑣 is a notation for translational speed or velocity. Yet, this problem involves rotational motion rather than translational. So we need to establish a bridge between rotational velocity and translational velocity.

Recall that 𝑣 represents our translational velocity. And this velocity is equal to π‘Ÿ, the radius of rotation multiplied by πœ”, the rotational velocity. It’s πœ” that we want to solve for in this problem represented by 𝑁, the number of revolutions per minute that this ride undergoes.

So let’s substitute π‘Ÿπœ” in for 𝑣 in our equation. We now have π‘Ÿπœ” quantity squared divided by π‘Ÿ is equal to the acceleration due to gravity. We can cancel out one of the factors of π‘Ÿ, leaving us with the equation π‘Ÿπœ” squared equals 𝑔.

Now let’s turn our attention to πœ”, the rotational speed of our ride. πœ” is equal to some number; we’ve called it 𝑁 of revolutions per minute. Let’s look a little bit into these units revolutions per minute.

One revolution around a circle is equal to two πœ‹ radians, so we can replace revolutions in this equation with two πœ‹ radians. And the denominator, one minute, is equal to 60 seconds. So πœ”, the rotational speed of a ride, is equal to some number, 𝑁, times two πœ‹ radians divided by 60 seconds. That’s the same as saying 𝑁 revolutions per minute.

Let’s substitute this expression for πœ” into our equation. We now get an equation that reads π‘Ÿ multiplied by 𝑁 two πœ‹ radians divided by 60 seconds quantity squared is equal to 𝑔, the acceleration due to gravity. We now want to rearrange this equation to solve for 𝑁. If we factor out an 𝑁 squared from inside the parentheses, our relationship becomes 𝑁 squared π‘Ÿ times the quantity two πœ‹ radians per 60 seconds squared equals 𝑔.

Let’s rearrange this equation to solve for 𝑁. First, we can divide both sides of the equation by π‘Ÿ, cancelling that term out on the left side of our equation. We can then multiply both sides of our equation by 60 seconds divided by two πœ‹ radians quantity squared, which then cancels that whole term out on the left side of our equation.

Finally, we take the square root of both sides of our resulting equation, which on the left side of our equation cancels out the square term above our 𝑁. Cleaning this up, we now have an equation for 𝑁. 𝑁 is equal to the square root of 𝑔 over π‘Ÿ multiplied by the square of 60 seconds divided by two πœ‹ radians.

When we plug in our values for 𝑔 and π‘Ÿ and enter these values into our calculator, we find a value of 𝑁 of 10.6 revolutions per minute. That’s how fast this ride must rotate in order for the riders to experience a centripetal acceleration equal in magnitude to the acceleration due to gravity.

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