Question

spin # 1
determine the stopping location of the prize wheel.
at this moment it is centered on the number 10. it is spinning at a rate of 104.90 rpm. it is slowing at a rate of 1.400 rad/s².
predict the time the wheel will remain spinning, the angular displacement it will go through, and then use this to predict the number on which the wheel will stop. to be safe, you will also get one number the right and to the left of the number you chose.
the wheel will spin clockwise.
when you are ready to enter your number, hit the end button.
end

Explanation:

Step1: Convert rpm to rad/s

First, we need to convert the initial angular velocity from rpm (revolutions per minute) to rad/s (radians per second). We know that 1 revolution is \(2\pi\) radians and 1 minute is 60 seconds. So the conversion formula is \(\omega_0=\frac{104.90\space rpm\times2\pi\space rad/rev}{60\space s/min}\).

Calculating that: \(\omega_0=\frac{104.90\times2\pi}{60}\approx104.90\times\frac{6.2832}{60}\approx\frac{660.1077}{60}\approx11.0018\space rad/s\)

Step2: Use angular kinematic equation

The wheel is slowing down, so it has an angular acceleration \(\alpha = - 1.400\space rad/s^2\) (negative because it's decelerating). We want to find the time \(t\) when the wheel stops, so the final angular velocity \(\omega = 0\space rad/s\).

The angular kinematic equation is \(\omega=\omega_0+\alpha t\). We can solve for \(t\):

\(t=\frac{\omega - \omega_0}{\alpha}\)

Substitute \(\omega = 0\), \(\omega_0\approx11.0018\space rad/s\), and \(\alpha=- 1.400\space rad/s^2\):

\(t=\frac{0 - 11.0018}{-1.400}=\frac{- 11.0018}{-1.400}\approx7.8584\space s\)

Step3: Calculate angular displacement

Now we use the angular displacement equation \(\theta=\omega_0t+\frac{1}{2}\alpha t^2\)

Substitute \(\omega_0\approx11.0018\space rad/s\), \(t\approx7.8584\space s\), and \(\alpha=- 1.400\space rad/s^2\):

\(\theta=(11.0018\times7.8584)+\frac{1}{2}\times(-1.400)\times(7.8584)^2\)

First term: \(11.0018\times7.8584\approx86.45\)

Second term: \(\frac{1}{2}\times(-1.400)\times61.75\approx - 0.7\times61.75\approx - 43.225\)

So \(\theta\approx86.45-43.225 = 43.225\space rad\)

Step4: Find the number of revolutions and the position

Each number on the wheel corresponds to an angular displacement of \(\Delta\theta=\frac{2\pi}{36}\space rad\) (since there are 36 numbers, so each number is \(10^\circ\) or \(\frac{2\pi}{36}\) radians per number). Wait, actually, looking at the wheel, there are 36 numbers (from 1 to 36? Wait, the wheel has numbers 1 - 36? Wait, the image shows numbers 1 - 36? Wait, the numbers on the wheel: looking at the image, the numbers go from 1 to 36? Wait, the pointer is at 10, and the wheel has 36 segments (since it's a prize wheel with 36 numbers? Let's check: from 1 to 36, so 36 numbers. So each number is \(\frac{2\pi}{36}=\frac{\pi}{18}\approx0.1745\space rad\) per number.

Wait, actually, when the wheel spins clockwise, the angular displacement \(\theta\) we calculated is the total angle rotated. Let's find how many full revolutions and the remaining angle.

First, find the number of full revolutions: \(n=\frac{\theta}{2\pi}=\frac{43.225}{6.2832}\approx6.88\) revolutions.

The fractional part is \(0.88\) revolutions. The angle for the fractional part is \(0.88\times2\pi\approx0.88\times6.2832\approx5.5292\space rad\)

Now, convert this fractional angle to the number of steps (each step is \(\frac{2\pi}{36}=\frac{\pi}{18}\approx0.1745\space rad\) per number). Wait, no: each number is a segment of \(\frac{2\pi}{36}\) radians. So the number of segments (numbers) moved from the initial position (10) is \(\frac{\theta}{\frac{2\pi}{36}}=\frac{\theta\times36}{2\pi}=\frac{43.225\times36}{6.2832}\approx\frac{1556.1}{6.2832}\approx247.66\)

Wait, the initial position is 10. Since the wheel spins clockwise, we add the number of segments (because clockwise rotation would move to higher numbers? Wait, no: if the wheel is at 10 and spins clockwise, the numbers increase? Wait, looking at the wheel: the numbers are arranged clockwise? Let's see: the pointer is at 10, and the numbers go clockwise: 10, 11, 12,... 36, 1, 2,... 9? Wait, no, the wheel in the image: the numbers are arranged such that 10 is at the pointer, and clockwise it goes 11,12,...36, then 1,2,...9? Wait, no, the wheel has numbers from 1 to 36, arranged clockwise, with 10 at the pointer. So when it spins clockwise, the wheel moves to higher numbers first, then wraps around.

Wait, maybe a better approach: the initial position is 10 (at angle 0, since it's centered on 10). The angular displacement \(\theta\) is clockwise, so the final position's angle is \(\theta\) (since we started at 10, angle 0, and rotated \(\theta\) clockwise). Now, each number is \(\frac{2\pi}{36}=\frac{\pi}{18}\) radians per number. So the number of numbers moved is \(\frac{\theta}{\frac{2\pi}{36}}=\frac{\theta\times36}{2\pi}\)

We have \(\theta\approx43.225\space rad\)

So \(\frac{43.225\times36}{2\pi}=\frac{1556.1}{6.2832}\approx247.66\) numbers.

Now, since the wheel is spinning clockwise, starting from 10, we add 247.66 to 10? Wait, no: wait, the initial position is 10. When the wheel spins clockwise, the numbers increase. Wait, no, if you have a wheel with numbers 1 - 36, and it's spinning clockwise, the pointer (which is fixed) will see the numbers moving counter-clockwise? Wait, no: if the wheel spins clockwise, the numbers move clockwise, so the pointer will see numbers decreasing? Wait, I think I made a mistake here.

Wait, let's clarify: when the wheel spins clockwise, the direction of rotation is clockwise, so the numbers on the wheel move clockwise. So the initial position is 10 (pointer at 10). After spinning clockwise, the wheel has rotated by \(\theta\) radians clockwise, so the new number at the pointer is 10 minus the number of numbers moved (because the wheel moved clockwise, so the numbers shift clockwise, so the pointer now points to a number that is 10 minus the number of steps, modulo 36). Wait, this is confusing. Let's use the angular displacement.

The total angular displacement is \(\theta\approx43.225\space rad\). Let's find the equivalent angle modulo \(2\pi\) (since a full revolution is \(2\pi\) radians). Wait, no: the wheel's numbers are arranged in 36 equal segments, so each segment is \(\frac{2\pi}{36}=\frac{\pi}{18}\approx0.1745\space rad\) per number. So the number of segments (numbers) is \(N=\frac{\theta}{\frac{2\pi}{36}}=\frac{\theta\times36}{2\pi}\)

Calculating \(N=\frac{43.225\times36}{6.2832}\approx\frac{1556.1}{6.2832}\approx247.66\)

Now, since the wheel is spinning clockwise, starting from 10, we need to find the final number. The formula for the final number (when spinning clockwise) is:

Final number \(= (10 + N)\space mod\space 36\)

Wait, no: if the wheel spins clockwise, the numbers move clockwise, so the pointer (fixed) will see the numbers moving in the clockwise direction, meaning the number at the pointer decreases? Wait, no, let's take a simple example: if the wheel is at 10 and spins clockwise by one segment (to number 11), the pointer would now point to 11? Wait, no, if the wheel spins clockwise, the number 11 moves to the pointer's position (which was at 10). So the pointer now points to 11. So spinning clockwise increases the number at the pointer.

So the initial number is 10. The number of segments moved is \(N\approx247.66\). So the final number is \(10 + 247.66=257.66\). Now, take modulo 36: \(257.66\div36 = 7\) with a remainder of \(257.66-7\times36=257.66 - 252 = 5.66\)

Wait, that can't be right. Wait, no, the angular displacement \(\theta\) is the total angle rotated by the wheel. So the wheel has rotated \(\theta\) radians clockwise. The initial position is 10 (angle 0 for the wheel's number 10). The wheel's number 10 is at angle 0, number 11 is at angle \(\frac{2\pi}{36}\), number 12 at \(\frac{4\pi}{36}\), etc. So the angle of number \(k\) on the wheel is \(\theta_k=(k - 10)\times\frac{2\pi}{36}\) (since number 10 is at 0). When the wheel rotates by \(\theta\) radians clockwise, the new angle of the wheel's number 10 is \(\theta\) (because the wheel rotated clockwise by \(\theta\)). So we want to find the number \(k\) such that \((k - 10)\times\frac{2\pi}{36}=\theta\space mod\space 2\pi\)

Wait, let's rephrase: the wheel's rotation is \(\theta\) radians clockwise. So the angle of the wheel's number 10 relative to the pointer is \(\theta\) radians clockwise. So the pointer is at angle 0 (fixed), and the wheel's number 10 is at angle \(\theta\) (clockwise). We need to find the number \(k\) on the wheel that is at angle 0 (pointer's position). So the angle of number \(k\) on the wheel is \((k - 10)\times\frac{2\pi}{36}\) (since number 10 is at 0, number 11 at \(\frac{2\pi}{36}\), etc.). We want this angle to be equal to \(-\theta\) (because the wheel rotated clockwise by \(\theta\), so the number \(k\) is \(\theta\) radians counter-clockwise from number 10? No, this is getting too confusing.

Alternative approach: each number is a segment of \(10^\circ\) (since 36 numbers, \(360^\circ/36 = 10^\circ\) per number). So each number is \(10^\circ\) or \(\frac{\pi}{18}\approx0.1745\space rad\) per number.

The total angular displacement \(\theta\approx43.225\space rad\). Convert \(\theta\) to degrees: \(\theta=43.225\times\frac{180}{\pi}\approx43.225\times57.2958\approx2476.6^\circ\)

Now, divide by \(360^\circ\) to find the number of full revolutions: \(2476.6\div360\approx6.88\) revolutions. The fractional part is \(0.88\times360\approx316.8^\circ\)

Now, since the wheel spins clockwise, the direction of rotation is clockwise, so the angle from the initial position (10) is \(316.8^\circ\) clockwise. Wait, but each number is \(10^\circ\), so the number of numbers moved is \(316.8\div10 = 31.68\) numbers.

Since the wheel spins clockwise, starting from 10, moving 31.68 numbers clockwise: 10 + 31.68 = 41.68. But since there are 36 numbers, we take modulo 36: 41.68 - 36 = 5.68. Wait, that's not right. Wait, no: if you move clockwise 31 numbers from 10, you get to 10 + 31 = 41, which is 41 - 36 = 5 (since 36 numbers, 41 is 5 in the next cycle). Then 0.68 numbers: 0.68\times10 = 6.8^\circ, so the number is 5 + 1 = 6? Wait, no, this is the mistake.

Wait, the initial position is 10. When the wheel spins clockwise, the numbers increase. So 10 -> 11 -> 12 ->... -> 36 -> 1 -> 2 ->... -> 9 -> 10. So each clockwise spin moves to the next higher number.

The total number of numbers moved is \(N=\frac{\theta}{\frac{2\pi}{36}}=\frac{43.225\times36}{2\pi}\approx\frac{1556.1}{6.2832}\approx247.66\) numbers.

Now, 247.66 divided by 36 (since there are 36 numbers) is 6 full cycles (6*36=216 numbers) and 247.66 - 216 = 31.66 numbers remaining.

So starting from 10, add 31.66 numbers: 10 + 31.66 = 41.66. Now, 41.66 - 36 = 5.66 (since 36 is the total numbers, so 41 is 5 in the next cycle). So the number is approximately 5.66, which is close to 6? Wait, no, maybe my initial assumption about the direction is wrong.

Wait, let's go back to the angular kinematic equation. Maybe I messed up the sign of the angular displacement. The wheel is decelerating, so \(\alpha\) is negative. The initial angular velocity is positive (clockwise? Wait, no: angular velocity direction: clockwise is negative in some conventions, but maybe I took it as positive. Let's redefine: let's take counter-clockwise as positive, so clockwise rotation is negative angular velocity.

Wait, maybe that's the mistake. Let's redefine:

Initial angular velocity: if the wheel is spinning clockwise, \(\omega_0=-11.0018\space rad/s\)

Angular acceleration: since it's slowing down (decelerating clockwise), \(\alpha = + 1.400\space rad/s^2\) (because if clockwise is negative, decelerating clockwise means acceleration is positive)

Then, the final angular velocity \(\omega = 0\)

Using \(\omega=\omega_0+\alpha t\)

\(0=-11.0018 + 1.400t\)

\(t=\frac{11.0018}{1.400}\approx7.8584\space s\) (same time as before)

Angular displacement \(\theta=\omega_0t+\frac{1}{2}\alpha t^2\)

\(\theta=(-11.0018\times7.8584)+\frac{1}{2}\times1.400\times(7.8584)^2\)

\(\theta=-86.45 + 0.7\times61.75\approx-86.45 + 43.225=-43.225\space rad\) (negative because clockwise)

Now, the magnitude of the angular displacement is \(43.225\space rad\),

Answer:

Step1: Convert rpm to rad/s

First, we need to convert the initial angular velocity from rpm (revolutions per minute) to rad/s (radians per second). We know that 1 revolution is \(2\pi\) radians and 1 minute is 60 seconds. So the conversion formula is \(\omega_0=\frac{104.90\space rpm\times2\pi\space rad/rev}{60\space s/min}\).

Calculating that: \(\omega_0=\frac{104.90\times2\pi}{60}\approx104.90\times\frac{6.2832}{60}\approx\frac{660.1077}{60}\approx11.0018\space rad/s\)

Step2: Use angular kinematic equation

The wheel is slowing down, so it has an angular acceleration \(\alpha = - 1.400\space rad/s^2\) (negative because it's decelerating). We want to find the time \(t\) when the wheel stops, so the final angular velocity \(\omega = 0\space rad/s\).

The angular kinematic equation is \(\omega=\omega_0+\alpha t\). We can solve for \(t\):

\(t=\frac{\omega - \omega_0}{\alpha}\)

Substitute \(\omega = 0\), \(\omega_0\approx11.0018\space rad/s\), and \(\alpha=- 1.400\space rad/s^2\):

\(t=\frac{0 - 11.0018}{-1.400}=\frac{- 11.0018}{-1.400}\approx7.8584\space s\)

Step3: Calculate angular displacement

Now we use the angular displacement equation \(\theta=\omega_0t+\frac{1}{2}\alpha t^2\)

Substitute \(\omega_0\approx11.0018\space rad/s\), \(t\approx7.8584\space s\), and \(\alpha=- 1.400\space rad/s^2\):

\(\theta=(11.0018\times7.8584)+\frac{1}{2}\times(-1.400)\times(7.8584)^2\)

First term: \(11.0018\times7.8584\approx86.45\)

Second term: \(\frac{1}{2}\times(-1.400)\times61.75\approx - 0.7\times61.75\approx - 43.225\)

So \(\theta\approx86.45-43.225 = 43.225\space rad\)

Step4: Find the number of revolutions and the position

Each number on the wheel corresponds to an angular displacement of \(\Delta\theta=\frac{2\pi}{36}\space rad\) (since there are 36 numbers, so each number is \(10^\circ\) or \(\frac{2\pi}{36}\) radians per number). Wait, actually, looking at the wheel, there are 36 numbers (from 1 to 36? Wait, the wheel has numbers 1 - 36? Wait, the image shows numbers 1 - 36? Wait, the numbers on the wheel: looking at the image, the numbers go from 1 to 36? Wait, the pointer is at 10, and the wheel has 36 segments (since it's a prize wheel with 36 numbers? Let's check: from 1 to 36, so 36 numbers. So each number is \(\frac{2\pi}{36}=\frac{\pi}{18}\approx0.1745\space rad\) per number.

Wait, actually, when the wheel spins clockwise, the angular displacement \(\theta\) we calculated is the total angle rotated. Let's find how many full revolutions and the remaining angle.

First, find the number of full revolutions: \(n=\frac{\theta}{2\pi}=\frac{43.225}{6.2832}\approx6.88\) revolutions.

The fractional part is \(0.88\) revolutions. The angle for the fractional part is \(0.88\times2\pi\approx0.88\times6.2832\approx5.5292\space rad\)

Now, convert this fractional angle to the number of steps (each step is \(\frac{2\pi}{36}=\frac{\pi}{18}\approx0.1745\space rad\) per number). Wait, no: each number is a segment of \(\frac{2\pi}{36}\) radians. So the number of segments (numbers) moved from the initial position (10) is \(\frac{\theta}{\frac{2\pi}{36}}=\frac{\theta\times36}{2\pi}=\frac{43.225\times36}{6.2832}\approx\frac{1556.1}{6.2832}\approx247.66\)

Wait, the initial position is 10. Since the wheel spins clockwise, we add the number of segments (because clockwise rotation would move to higher numbers? Wait, no: if the wheel is at 10 and spins clockwise, the numbers increase? Wait, looking at the wheel: the numbers are arranged clockwise? Let's see: the pointer is at 10, and the numbers go clockwise: 10, 11, 12,... 36, 1, 2,... 9? Wait, no, the wheel in the image: the numbers are arranged such that 10 is at the pointer, and clockwise it goes 11,12,...36, then 1,2,...9? Wait, no, the wheel has numbers from 1 to 36, arranged clockwise, with 10 at the pointer. So when it spins clockwise, the wheel moves to higher numbers first, then wraps around.

Wait, maybe a better approach: the initial position is 10 (at angle 0, since it's centered on 10). The angular displacement \(\theta\) is clockwise, so the final position's angle is \(\theta\) (since we started at 10, angle 0, and rotated \(\theta\) clockwise). Now, each number is \(\frac{2\pi}{36}=\frac{\pi}{18}\) radians per number. So the number of numbers moved is \(\frac{\theta}{\frac{2\pi}{36}}=\frac{\theta\times36}{2\pi}\)

We have \(\theta\approx43.225\space rad\)

So \(\frac{43.225\times36}{2\pi}=\frac{1556.1}{6.2832}\approx247.66\) numbers.

Now, since the wheel is spinning clockwise, starting from 10, we add 247.66 to 10? Wait, no: wait, the initial position is 10. When the wheel spins clockwise, the numbers increase. Wait, no, if you have a wheel with numbers 1 - 36, and it's spinning clockwise, the pointer (which is fixed) will see the numbers moving counter-clockwise? Wait, no: if the wheel spins clockwise, the numbers move clockwise, so the pointer will see numbers decreasing? Wait, I think I made a mistake here.

Wait, let's clarify: when the wheel spins clockwise, the direction of rotation is clockwise, so the numbers on the wheel move clockwise. So the initial position is 10 (pointer at 10). After spinning clockwise, the wheel has rotated by \(\theta\) radians clockwise, so the new number at the pointer is 10 minus the number of numbers moved (because the wheel moved clockwise, so the numbers shift clockwise, so the pointer now points to a number that is 10 minus the number of steps, modulo 36). Wait, this is confusing. Let's use the angular displacement.

The total angular displacement is \(\theta\approx43.225\space rad\). Let's find the equivalent angle modulo \(2\pi\) (since a full revolution is \(2\pi\) radians). Wait, no: the wheel's numbers are arranged in 36 equal segments, so each segment is \(\frac{2\pi}{36}=\frac{\pi}{18}\approx0.1745\space rad\) per number. So the number of segments (numbers) is \(N=\frac{\theta}{\frac{2\pi}{36}}=\frac{\theta\times36}{2\pi}\)

Calculating \(N=\frac{43.225\times36}{6.2832}\approx\frac{1556.1}{6.2832}\approx247.66\)

Now, since the wheel is spinning clockwise, starting from 10, we need to find the final number. The formula for the final number (when spinning clockwise) is:

Final number \(= (10 + N)\space mod\space 36\)

Wait, no: if the wheel spins clockwise, the numbers move clockwise, so the pointer (fixed) will see the numbers moving in the clockwise direction, meaning the number at the pointer decreases? Wait, no, let's take a simple example: if the wheel is at 10 and spins clockwise by one segment (to number 11), the pointer would now point to 11? Wait, no, if the wheel spins clockwise, the number 11 moves to the pointer's position (which was at 10). So the pointer now points to 11. So spinning clockwise increases the number at the pointer.

So the initial number is 10. The number of segments moved is \(N\approx247.66\). So the final number is \(10 + 247.66=257.66\). Now, take modulo 36: \(257.66\div36 = 7\) with a remainder of \(257.66-7\times36=257.66 - 252 = 5.66\)

Wait, that can't be right. Wait, no, the angular displacement \(\theta\) is the total angle rotated by the wheel. So the wheel has rotated \(\theta\) radians clockwise. The initial position is 10 (angle 0 for the wheel's number 10). The wheel's number 10 is at angle 0, number 11 is at angle \(\frac{2\pi}{36}\), number 12 at \(\frac{4\pi}{36}\), etc. So the angle of number \(k\) on the wheel is \(\theta_k=(k - 10)\times\frac{2\pi}{36}\) (since number 10 is at 0). When the wheel rotates by \(\theta\) radians clockwise, the new angle of the wheel's number 10 is \(\theta\) (because the wheel rotated clockwise by \(\theta\)). So we want to find the number \(k\) such that \((k - 10)\times\frac{2\pi}{36}=\theta\space mod\space 2\pi\)

Wait, let's rephrase: the wheel's rotation is \(\theta\) radians clockwise. So the angle of the wheel's number 10 relative to the pointer is \(\theta\) radians clockwise. So the pointer is at angle 0 (fixed), and the wheel's number 10 is at angle \(\theta\) (clockwise). We need to find the number \(k\) on the wheel that is at angle 0 (pointer's position). So the angle of number \(k\) on the wheel is \((k - 10)\times\frac{2\pi}{36}\) (since number 10 is at 0, number 11 at \(\frac{2\pi}{36}\), etc.). We want this angle to be equal to \(-\theta\) (because the wheel rotated clockwise by \(\theta\), so the number \(k\) is \(\theta\) radians counter-clockwise from number 10? No, this is getting too confusing.

Alternative approach: each number is a segment of \(10^\circ\) (since 36 numbers, \(360^\circ/36 = 10^\circ\) per number). So each number is \(10^\circ\) or \(\frac{\pi}{18}\approx0.1745\space rad\) per number.

The total angular displacement \(\theta\approx43.225\space rad\). Convert \(\theta\) to degrees: \(\theta=43.225\times\frac{180}{\pi}\approx43.225\times57.2958\approx2476.6^\circ\)

Now, divide by \(360^\circ\) to find the number of full revolutions: \(2476.6\div360\approx6.88\) revolutions. The fractional part is \(0.88\times360\approx316.8^\circ\)

Now, since the wheel spins clockwise, the direction of rotation is clockwise, so the angle from the initial position (10) is \(316.8^\circ\) clockwise. Wait, but each number is \(10^\circ\), so the number of numbers moved is \(316.8\div10 = 31.68\) numbers.

Since the wheel spins clockwise, starting from 10, moving 31.68 numbers clockwise: 10 + 31.68 = 41.68. But since there are 36 numbers, we take modulo 36: 41.68 - 36 = 5.68. Wait, that's not right. Wait, no: if you move clockwise 31 numbers from 10, you get to 10 + 31 = 41, which is 41 - 36 = 5 (since 36 numbers, 41 is 5 in the next cycle). Then 0.68 numbers: 0.68\times10 = 6.8^\circ, so the number is 5 + 1 = 6? Wait, no, this is the mistake.

Wait, the initial position is 10. When the wheel spins clockwise, the numbers increase. So 10 -> 11 -> 12 ->... -> 36 -> 1 -> 2 ->... -> 9 -> 10. So each clockwise spin moves to the next higher number.

The total number of numbers moved is \(N=\frac{\theta}{\frac{2\pi}{36}}=\frac{43.225\times36}{2\pi}\approx\frac{1556.1}{6.2832}\approx247.66\) numbers.

Now, 247.66 divided by 36 (since there are 36 numbers) is 6 full cycles (6*36=216 numbers) and 247.66 - 216 = 31.66 numbers remaining.

So starting from 10, add 31.66 numbers: 10 + 31.66 = 41.66. Now, 41.66 - 36 = 5.66 (since 36 is the total numbers, so 41 is 5 in the next cycle). So the number is approximately 5.66, which is close to 6? Wait, no, maybe my initial assumption about the direction is wrong.

Wait, let's go back to the angular kinematic equation. Maybe I messed up the sign of the angular displacement. The wheel is decelerating, so \(\alpha\) is negative. The initial angular velocity is positive (clockwise? Wait, no: angular velocity direction: clockwise is negative in some conventions, but maybe I took it as positive. Let's redefine: let's take counter-clockwise as positive, so clockwise rotation is negative angular velocity.

Wait, maybe that's the mistake. Let's redefine:

Initial angular velocity: if the wheel is spinning clockwise, \(\omega_0=-11.0018\space rad/s\)

Angular acceleration: since it's slowing down (decelerating clockwise), \(\alpha = + 1.400\space rad/s^2\) (because if clockwise is negative, decelerating clockwise means acceleration is positive)

Then, the final angular velocity \(\omega = 0\)

Using \(\omega=\omega_0+\alpha t\)

\(0=-11.0018 + 1.400t\)

\(t=\frac{11.0018}{1.400}\approx7.8584\space s\) (same time as before)

Angular displacement \(\theta=\omega_0t+\frac{1}{2}\alpha t^2\)

\(\theta=(-11.0018\times7.8584)+\frac{1}{2}\times1.400\times(7.8584)^2\)

\(\theta=-86.45 + 0.7\times61.75\approx-86.45 + 43.225=-43.225\space rad\) (negative because clockwise)

Now, the magnitude of the angular displacement is \(43.225\space rad\),