# Video: Using Kinematics Equations to Calculate the Angular Acceleration

A phonograph turntable rotating at 33 1/3 rpm slows down and stops in 1.00 min. What is the turntable’s angular acceleration assuming it is constant? How many complete revolutions does the turntable make while stopping?

04:09

### 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.