Question

Mary is an avid game show fan and one of the contestants on a popular game show. She spins the wheel, and after 5.5 revolutions, the wheel comes to rest on a space that has a $1500 value prize. If the initial angular speed of the wheel is 3.80 rad/s, find the angle through which the wheel has turned when the angular speed reaches 2.40 rad/s.

Answer 1

θ = 3.306 * π

Step-by-step explanation:

Convert 1.5 revolutions to radians.

1 revolution = 2π radians.

5.5 revolutions = 11π radians

Determine the wheel’s angular acceleration.

ωf^2 = ωi^2 + 2 * α * θ, ωf

ωf = 0

0 = 3.8^2 + 2 * α * 11π

-14.44 = 22 * π * α

α = -14.44 ÷ (11 * π)

α = -0.656 rad/s^2

In the same equation to determine the angle.

ωf^2 = ωi^2 + 2 * α * θ, ωf = 0

2.4^2 = 3.8^2 + 2 * [-14.44 ÷ (11 * π)] * θ

2.4^2 = 3.8^2 – [28.88 ÷ (11 * π)] * θ

-8.68 = -28.88 ÷ (11 * π)* θ

-95.48* π = -28.88* θ

θ = -95.48*π ÷ -28.88

θ = 3.306 * π

Answer 2

`6.58\pi rad` or
`1183.64^o`

Step-by-step explanation:

Let the initial angular velocity be
`\omega_o` and the angular deceleration be
`\alpha`. Equation (1) is the equivalent of the third equation of motion for a uniformly accelerated or decelerate linear motion for circular or angular motion, since the radius of the motion specified by the problem does not change.

`\omega_1^2=\omega_o^2+2\alpha\theta.................(1)`

where
`\omega_1` is the final angular velocity and
`\theta` is the angle turned through.

Also equation (2) is a relationship between angular velocity
`\theta` and number of revolution n;

`\theta=2\pi n.................(2)`

Substituting equation (2) into (1) we obtain the following;

`\omega_1^2=\omega_o^2+4\alpha \pi n.............. (3)`

In the problem, our first mission is find the value of
`\alpha`. When Mary spins the wheel with an initial angular velocity of 3.8rad/s, it comes to rest after making 5.5 revolutions. Therefore;

`\omega_o=3.8rad/s\\n=5.5\\\omega_1=0m/s`

It should be noted that the wheel comes to rest that is why the final angular velocity
`\omega_1=0`

The angular acceleration is obtained by making
`\alpha` the subject of formula from equation (3)

`\alpha =(\omega_1^2-\omega_o^2)/(4\pi n)..............(4)`

hence;

`\alpha =(0^2-3.8^2)/(4\pi *5.5) =(14.44)/(22\pi )\\`

`\alpha =(-0.66)/(\pi ) rad/s^2`

The negative result is an indication that the wheel is actually decelerating.

We then use equation (1) to determine the angle turned through when its angular velocity is 2.4rad/s as follows by making
`\theta` the subject of formula.

`\theta=(\omega_1^2-\omega_2^2)/(2\alpha)............(5)`

In this case,
`\omega_1=2.4rad/s`

Therefore;

`\theta=(2.4^2-3.8^2)/(2((-0.66)/(\pi)))\\\\\\\theta=(-8.68\pi)/(-1.32)\\\\\theta=6.58\pi rad`

`Recall\\\pi rad=180^o\\hence\\6.58\pi=6.58*180^o=1183.64^o`