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 π΄ to a point π΅. The ratio of the magnitude of the centripetal acceleration at point π΄ to the magnitude of the centripetal acceleration at point π· is πβ. The ratio of the magnitude of the centripetal acceleration at point π΄ to the magnitude of the centripetal acceleration at point π΅ is πβ. What is πβ divided by πβ?

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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 π΄ to a point π΅. The ratio of the magnitude of the centripetal acceleration at point π΄ to the magnitude of the centripetal acceleration at point π· is π one. The ratio of the magnitude of the centripetal acceleration at point π΄ to the magnitude of the centripetal acceleration at point π΅ is π two. What is π one divided by π two?

This question refers to a set of points on an object that moves in a circular paths. These points comprise the pair of point π΄ and point π΅ and the pair of point π΄ and point π·. Points π΄ and π΅ are on the same circular path, while points π΄ and π· are on different ones. When a part of the rope is at any of these points, it accelerates centripetally toward the center of the circle. The ratio of the ropeβs centripetal acceleration at point π΄ to that at point π· is π one. The ratio is π two for the centripetal acceleration at point π΄ over that at point π΅. We want to solve for π one divided by π two.

Letβs first remember that centripetal acceleration can be expressed as π equals π£ squared over π, where π£ is the objectβs tangential velocity a distance π from the center of the circular arc the object travels in. We can also write this expression as π΄ equals Ο squared times π, where Ο is the angular velocity expressed in radians per second. As the rope in this example rotates, at some points it rotates faster and another points more slowly. At any given position though, the angular speed of the rope is the same for all points along its length. So for example, the ropeβs angular speed at point π΄ is the same as its angular speed at point πΆ.

Our question statement tells us that the free end of the rope moves at constant speed from point π΄ to point π΅. This means the angular speeds of the rope at these points, weβll call those speeds Ο π΄ and Ο π΅, are equal. Since points π΄ and π΅ are also the same distance from the fixed end of the rope, by our earlier equation, we can say that the ropeβs centripetal acceleration at point π΄, weβll call it π sub π΄, is equal to the ropeβs centripetal acceleration at point π΅, π sub π΅. This means that π two must equal one.

Next, letβs consider points π΄ and π·. Letβs call the distance from point π΄ to the fixed end of the rope π sub π΄, thatβs 0.22 meters, and the distance from point π· to the fixed end π sub π·. Thatβs 0.16 meters. Weβre told that the rope takes 0.65 seconds to complete a full rotation. Knowing this and recalling that speed equals distance divided by time, we can solve for the tangential speed of the rope at points π΄ and π·. Clearing some space on screen, we can write that the ropeβs tangential speed at point π΄, π£ sub π΄, is two times π times π sub π΄ divided by 0.65 seconds.

Recalling our first equation for centripetal acceleration, we can say the ropeβs centripetal acceleration at point π΄ equals π£ sub π΄ squared divided by π sub π΄. Since π£ sub π΄ equals two times π times π sub π΄ over 0.65 seconds, we know that one factor of π sub π΄ will cancel from numerator and denominator. π sub π΄ equals four π squared times π sub π΄ divided by 0.65 seconds squared. π sub π΄ is 0.22 meters, so π sub π΄ is 20.556 and so on meters per second squared.

Storing this result off to the side, weβll now do a similar calculation for π sub π·. The tangential velocity of the rope at point π·, π£ sub π·, equals two times π times π sub π· divided by 0.65 seconds. π sub π· is π£ sub π· squared over π sub π·. And plugging in for π£ sub π·, we again find that a factor of the radial distance cancels from top and bottom. π sub π· equals four π squared times π sub π· divided by 0.65 seconds squared. π sub π· is 0.16 meters. So π sub π· calculates out to 14.950 and so on meters per second squared.

Now that we know both π sub π΄ and π sub π·, we can solve for their ratio. To three decimal places, it is 1.375. Note that the units have canceled out from the expression entirely. Our question asked us to solve for the ratio π one to π two. Since π two is just one, the ratio equals π one itself, or 1.375. This is the ratio of the centripetal acceleration of the rope at point π΄ over that at point π· to the centripetal acceleration of the rope at point π΄ to that at point π΅.

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