Video Transcript
A helicopter flies in a circle of
radius 375 meters, taking a time of 42 seconds to complete one revolution. What is the magnitude of the
acceleration of the helicopter toward the center of its circular path?
Let’s start by drawing a
diagram. Here’s the helicopter flying in its
circular path with a radius, which we’ll call 𝑟, of 375 meters. Now, because the helicopter is just
flying in a circle, we can describe its motion using angular velocity represented by
𝜔. Recall that angular velocity is
defined as a change in angular displacement, or Δ𝜃, divided by a change in time, or
Δ𝑡. Now, the question statement told us
a time interval of 42 seconds, which is how long it takes the helicopter to complete
one full revolution around its circular path. We’ll use this as our Δ𝑡
value.
Over that time span, the
helicopter’s change in angular displacement Δ𝜃 is one revolution or one full turn
around a circle. But remember that when we’re
calculating, we should describe angular displacement using radians. And one full revolution is equal to
two 𝜋 radians. Now substituting both of these
values into the formula for angular velocity, we have 𝜔 equals two 𝜋 radians
divided by 42 seconds. And since 42 equals two times 21,
we can cancel out a factor of two from the numerator and denominator, thus
simplifying the fraction to 𝜋 radians divided by 21 seconds or 𝜋 over 21 radians
per second.
Now, this question is asking us to
find the magnitude of the acceleration that the helicopter experiences in the
direction of the center of its circular path. To better understand this, recall
that any object in uniform circular motion, like the helicopter here, must be
constantly accelerating toward the center of its circular path. This acceleration is known as
centripetal acceleration, represented by 𝑎 subscript c. Further, recall that given an
object’s angular velocity 𝜔 and the radius 𝑟 of its circular path, we can
calculate its centripetal acceleration using the formula 𝜔 squared times 𝑟.
Now, we already have values for
both 𝜔 and 𝑟 expressed in base SI units. So, let’s copy the formula and
substitute them in. This gives us 𝜋 over 21 radians
per second squared times 375 meters. Before we start calculating though,
let’s take a moment to think about the units here. We need to recall that while
angular units such as radians, degrees, and revolutions do help us keep track of
angular displacement, they’re technically dimensionless. They don’t have a physical element
associated with them unlike other units we might be more familiar with, such as
meters or seconds.
But this does not mean that all
angular units are equivalent. It still very much matters which
one we use. Physics equations were actually
designed with radians in mind. And thus for calculations, like
this one here, we do need to express angular velocity using radians per second. It’s just that when it comes to the
physical dimensions of angular velocity, we’re only concerned with the per second
part.
Okay, now back to the units for
centripetal acceleration. Squaring angular velocity will give
us per second squared, and of course radius contributes a factor of meters. Together, we’ll have units of
meters per second squared, which are the proper SI units for acceleration. Now entering this into a
calculator, 𝜔 squared times 𝑟 comes out to 8.393 and so on meters per second
squared. And finally, rounding this value to
one decimal place, we found that the magnitude of the acceleration of the helicopter
toward the center of its circular path is 8.4 meters per second squared.
