Video Transcript
Two cases of the rotational motion
of an object are shown in the following diagram. The position of the object at three
instants is shown. The interval 𝑡 one minus 𝑡 zero
equals the interval 𝑡 two minus 𝑡 one. In both cases, △𝜃 one equals △𝜃
two. Which of the following most
correctly compares the rotational motion occurring in the two cases? (A) Rotational motion I shows a
greater constant angular speed than rotational motion II. (B) Rotational motion I shows
angular acceleration, and rotational motion II shows constant angular speed. (C) Rotational motion I shows
constant angular speed, and rotational motion II shows angular acceleration. (D) Both cases show the same value
of constant angular speed.
Let’s consider these two examples
of rotational motion of an object. Right away, we can see that the
radius of the circle in rotational motion I is larger than the radius of the circle
in rotational motion II. Nonetheless, notice that at what we
can call 12 o’clock on the circle is where the time 𝑡 zero occurs. And as we think about the object’s
motion during the time interval from 𝑡 zero up to 𝑡 one, we see that in both cases
the object moves through an angle of △𝜃 one. In other words, in each case, over
the same time interval, the angular displacement of the object is the same.
Now let’s consider the time 𝑡 one
in both instants and the interval of time from 𝑡 one up to 𝑡 two. Once again, the change in angular
position of our object in both cases is the same over this time interval. This time, that change is called
△𝜃 two. Along with all this, we see that in
both rotational motion I and rotational motion II, at the time 𝑡 zero, the object
is moving at a speed labeled as 𝑣 zero. Then once again in both cases, at
𝑡 one, the object is moving at a speed 𝑣 one and at 𝑡 two, it’s moving at a speed
𝑣 two, again, in both types of rotational motion.
Our problem statement tells us that
the first interval of time, from 𝑡 zero to 𝑡 one, is equal to the second interval,
from 𝑡 one to 𝑡 two. Along with this, it tells us that
the first change in angular position, △𝜃 one, equals the second change △𝜃
two.
Let’s recall at this point the
definition of angular speed. Angular speed, 𝜔, is equal to a
change in angular position, △𝜃, divided by a change in time, △𝑡. If we were to calculate the angular
speed of our object in rotational motion I over the first time interval, and we can
call this angular speed 𝜔 one, that will be equal, we see, to the change in angular
position, △𝜃 one, divided by the change in time, which is 𝑡 one minus 𝑡
zero. If we were to follow the same
process but for the second time interval, from 𝑡 one to 𝑡 two, then we could call
that angular speed 𝜔 two. 𝜔 two is equal to △𝜃 two divided
by 𝑡 two minus 𝑡 one.
Let’s recall though that △𝜃 one
equals △𝜃 two. And 𝑡 one minus 𝑡 zero equals 𝑡
two minus 𝑡 one. That is, for rotational motion I,
𝜔 one equals 𝜔 two. This tells us that the angular
speed of our object is constant throughout its motion. Let’s now remember that angular
acceleration, 𝛼, is equal to a change in angular speed, △𝜔, divided by a change
in time, △𝑡. We’ve just seen that in the case of
rotational motion I, we have no change in angular speed. In other words, △𝜔 for rotational
motion I is zero. Therefore, the object in this case
has no angular acceleration. This allows us to eliminate one of
our answer options. Answer option (B) says that
rotational motion I shows angular acceleration. We know this is not the case,
rather that rotational motion I shows constant angular speed. So we can eliminate this
option.
Clearing that away, let’s now
consider the object in rotational motion II. Just like before, we can write
expressions for the angular speed of our object in rotational motion two over the
first time interval and over the second. Also just like before, 𝜔 one, the
angular speed of our object over that first time interval, equals △𝜃 one divided
by 𝑡 one minus 𝑡 zero. Likewise, 𝜔 two equals △𝜃 two
divided by 𝑡 two minus 𝑡 one. We found then that 𝜔 one equals 𝜔
two for rotational motion II. And these quantities are both the
same as they were in rotational motion I. We can say then that in neither one
of our rotational motion cases is the object accelerating. Notice that that allows us to
eliminate answer option (C).
We can also eliminate answer option
(A), which says that while both cases show us constant angular speed, one of these
speeds is greater than the other. We now know that that’s not true,
that actually in both cases the constant angular speed is the same. Answer option (D) accurately
describes what’s really going on. We may wonder though, just what is
the difference between the motion of the object in rotational motion one and that in
rotational motion II. We can recall that for an object
experiencing constant angular speed, moving through a circle with a radius 𝑟, the
product of that circle’s radius and its constant angular speed, 𝜔, is equal to its
linear speed, 𝑣. In both of our instances, 𝜔, the
constant angular speed, is the same.
We saw at the outset, though, that
the radius of motion in rotational motion I is greater than that in rotational
motion II. This means that even though both of
our objects will move in a constant linear speed, the linear speed of the object in
rotational motion I will be greater than that in rotational motion II. Note that this is somewhat similar
to our answer option (A). The reason that option was
incorrect though is it specifies rotational motion. It’s only the case for linear
motion that the linear speed of the object in rotational motion I is greater than
that in rotational motion II. All this to say, the correct answer
to our question is option (D): both cases show the same value of constant angular
speed.
