Question Video: Comparing Centripetal Accelerations at Different Points | Nagwa Question Video: Comparing Centripetal Accelerations at Different Points | Nagwa

Question Video: Comparing Centripetal Accelerations at Different Points Physics • First Year of Secondary School

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A uniform rope is rotated horizontally around one of its ends, as shown in the diagram. The end of the rope opposite to the fixed end returns to its position every 0.65 s. The free end of the rope moves at constant speed from a point 𝐴 to a point 𝐡. What is the ratio of the magnitude of the centripetal acceleration of the point 𝐴 to the magnitude of the centripetal acceleration at point 𝐡? What is the ratio of the magnitude of the centripetal acceleration of the point 𝐴 to the magnitude of the centripetal acceleration at point 𝐷?

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Video Transcript

A uniform rope is rotated horizontally around one of its ends, as shown in the diagram. The end of the rope opposite to the fixed end returns to its position every 0.65 seconds. The free end of the rope moves at constant speed from a point 𝐴 to a point 𝐡. What is the ratio of the magnitude of the centripetal acceleration of the point 𝐴 to the magnitude of the centripetal acceleration at point 𝐡?

In our diagram, we see this rope at two different positions. The rope is rotated so that it moves in a circle with its fixed end located here. As it turns, the free end of the rope passes through these points 𝐡 and 𝐴. Both of these points are at the same radial distance 0.22 meters from the fixed end of the rope.

To answer this first part of our question that talks about centripetal acceleration, let’s start by noting that if we have an object that’s moving in a circular path, this blue object here, and if the object moves at a linear speed 𝑣, moving around a circle of radius π‘Ÿ, then that object’s centripetal acceleration, we’ll call it π‘Ž sub c, is equal to 𝑣 squared divided by π‘Ÿ. This is the acceleration the object experiences towards the center of its circular arc. Indeed, the word β€œcentripetal” means center seeking. Any time any object moves in a circular path, for example, the end of our rope, it experiences centripetal acceleration. We want to compare the centripetal acceleration of the rope as it’s at point 𝐴 to that as it’s at point 𝐡.

In our problem statement, we’re told that the free end of the rope moves at constant speed between these two points. We can write then that the speed of the rope as it passes through point 𝐴 is equal to the speed of the rope as it passes through point 𝐡. Since these speeds are equal, we’ll call them simply 𝑣. We can therefore write the centripetal acceleration of the end of the rope at point 𝐴 as 𝑣 squared divided by π‘Ÿ sub 𝐴, where π‘Ÿ sub 𝐴 is the distance between the fixed end of the rope and point 𝐴. Likewise, the centripetal acceleration of the rope as it passes point 𝐡 is 𝑣 squared over π‘Ÿ sub 𝐡.

We saw earlier though that the distance between the fixed end of the rope and points 𝐴 and 𝐡 is the same. We can write then that π‘Ÿ sub 𝐴 equals π‘Ÿ sub 𝐡. And with these values being the same, we can call them simply π‘Ÿ. Using these symbols, we see that the centripetal acceleration of the rope at point 𝐴 is the same as it is at point 𝐡. This means that when we find the ratio of the magnitude of the centripetal acceleration at point 𝐴 to that at point 𝐡, that ratio is one. This must be so because the rope’s centripetal acceleration at both points 𝐴 and 𝐡 is the same.

Let’s now clear some space on screen and look at part two of our question.

What is the ratio of the magnitude of the centripetal acceleration of the point 𝐴 to the magnitude of the centripetal acceleration at point 𝐷?

We see that point 𝐷 is a bit closer in towards the fixed end of the rope than points 𝐴 or 𝐡. Earlier, we wrote that the magnitude of the centripetal acceleration of the rope at point 𝐴 equals 𝑣 squared over π‘Ÿ sub 𝐴. Now, to make this expression completely specific to point 𝐴, let’s call the speed 𝑣 sub 𝐴. So, when the end of our rope is at point 𝐴, that end moves with a speed 𝑣 sub 𝐴, and it’s a distance π‘Ÿ sub 𝐴 from the fixed end of the rope. For the part of the rope that passes through point 𝐷 though, that point will move with a speed 𝑣 sub 𝐷. And it will be a distance π‘Ÿ sub 𝐷 away from the fixed end of the rope.

At this point, let’s recall the relationship for the linear speed 𝑣 of an object that moves in a circle to its angular speed πœ”. When the radius of the circle that the object moves through is π‘Ÿ, 𝑣 is equal to π‘Ÿ times πœ”. We bring this up because it will help us write the centripetal acceleration of an object in a different way. If we replace 𝑣 with π‘Ÿ times πœ”, then the equation for centripetal acceleration becomes π‘Ÿ squared πœ” squared over π‘Ÿ or, since one factor of π‘Ÿ cancels from numerator and denominator, π‘Ÿ times πœ” squared.

As we think about our rope rotating about one of its ends, we can realize that the angular speed of any point along the rope is the same. That is, it takes the same amount of time for a point, say here on the rope, to go all the way around the circle once as it takes for a point, say here or here or at any other point on the rope, to complete one full revolution. The angular speed πœ” of a point on the rope is independent of that point’s distance from the fixed end of the rope.

This all means that we can rewrite our expressions for the centripetal acceleration of the rope at points 𝐴 and 𝐷. Instead of using the linear speeds 𝑣 sub 𝐴 and 𝑣 sub 𝐷, we can express centripetal acceleration in terms of the angular speed πœ”, which is the same at both points. When we consider then the ratio that our question asks about, this is equal to π‘Ÿ sub 𝐴 times πœ” squared divided by π‘Ÿ sub 𝐷 times πœ” squared. Note that the angular speed πœ” cancels from numerator and denominator. The ratio of centripetal acceleration at these two points then simply equals the ratio of these radii, π‘Ÿ sub 𝐴 and π‘Ÿ sub 𝐷.

Our diagram shows us that π‘Ÿ sub 𝐴 is 0.22 meters. π‘Ÿ sub 𝐷 is 0.16 meters. In this fraction, the units of meters will cancel from numerator and denominator. 0.22 divided by 0.16 is equal to exactly 1.375. This is the ratio of the magnitude of the centripetal acceleration of the point 𝐴 to that at point 𝐷.

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