Video Transcript
A phonograph turntable rotating at
33 and a third rpm slows down and stops in 1.00 minutes. What is the turntable’s angular
acceleration assuming it is constant? How many complete revolutions does
the turntable make while stopping?
We’ll call the turntable’s angular
acceleration 𝛼 and the number of complete revolutions it makes while stopping
𝑁. In this scenario, we have a
rotating turntable which initially has an angular speed we’ve called capital Ω, 33
to third revolutions per minute. Over a time span of 1.00 minutes,
it slows to a stop. Since we’re told that its angular
acceleration as it does so is constant, that means that the kinematic equations of
motion apply to the motion of the turntable. In particular, we’re helped by the
equation which says that 𝜔 sub 𝑓, the final angular speed, is equal to 𝜔 sub
zero, the initial angular speed, plus 𝛼, the angular acceleration, times time.
In this expression, our angular
speeds are written in units of radians per second. When we apply this relationship to
our scenario, we know that because the turntable ends up at rest, 𝜔 sub 𝑓 is
zero. And we want now to solve for 𝜔 sub
zero, the initial angular speed of our disc, in units of radians per second. To do that, let’s start by writing
down capital Ω. If we multiply capital Ω by the
conversion factor two 𝜋 radians are in every revolution, then we’ve effectively
changed the units of this expression now to radians per minute. To get the units to radians per
second, we can multiply by the conversion factor one minute is in every 60 seconds,
cancelling out the units of minutes. We now have our expression for 𝜔
sub zero. It’s 33 and a third times two 𝜋
over 60 radians per second.
With that value known, we can now
rearrange our kinematic equation to solve for the angular acceleration 𝛼. 𝛼 is equal to negative 𝜔 sub zero
over 𝑡. Plugging in these values, we see 𝛼
is negative 33.33 repeating 𝜋 over 30 times one over 60 in units of radians per
second squared. This is equal to negative 0.058
radians per second squared. That’s the angular acceleration of
the disk as it slows down.
Next, we wanna solve for the number
of complete rotations the phonograph goes through as it slows to a stop. Considering what we know: we know
the initial angular speed of the phonograph. We know its angular
acceleration. We know its final angular speed
zero. And we essentially wanna calculate
an angular distance that is travelled. We can begin solving for 𝑁 by
working from a kinematic equation that lets us solve for the angular displacement
𝜃. The total angular displacement of
the phonograph as it comes to a stop is 𝜃. And if we divide 𝜃 by two 𝜋
radians, then that will equal a total number of revolutions 𝑁 we complete.
Since we’ve solved for 𝜔 sub zero
and 𝛼 and we’re given 𝑡 in the problem statement, we’re ready to plug in and solve
for 𝑁. When we have our values for 𝜔 sub
zero and 𝛼 and 𝑡 in units of seconds plugged into this equation, notice as we go
from left to right that the units of seconds cancel out in our first term. The units of seconds squared cancel
out in our second term. And because we’re dividing all of
this by two 𝜋 radians, the units of radians also cancel out, leaving us with a
unitless result. When we enter this expression on
our calculator, we get a result of approximately 16.7. But since 𝑁 is the number of
complete revolutions the phonograph makes before it comes to a stop, that means that
𝑁 is 16. And so, we’ve used the rotational
kinematic equations to solve for the number of times the phonograph spins before
coming to a stop.
