### 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 ropeβs 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.

Letβs first remember that
centripetal acceleration can be expressed as π equals π£ squared over π, where π£
is the objectβs 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 ropeβs angular
speed at point π΄ is the same as its angular speed at point πΆ.

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, weβll 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 ropeβs centripetal acceleration at point π΄, weβll call it π sub π΄, is
equal to the ropeβs centripetal acceleration at point π΅, π sub π΅. This means that π
two must equal
one.

Next, letβs consider points π΄ and
π·. Letβs call the distance from point
π΄ to the fixed end of the rope π sub π΄, thatβs 0.22 meters, and the distance from
point π· to the fixed end π sub π·. Thatβs 0.16 meters. Weβre 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 ropeβs 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 ropeβs 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, weβll 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΅.