Wheel on a game show force problem

In summary, the conversation discusses finding the average torque exerted on a wheel on a game show. The wheel has an initial angular speed of 1.3 rad/s and comes to rest after rotating through 3/4 of a turn. The wheel is a disk with a radius of 0.80m and a mass of 6.4 kg. The person used the equation w^2=w(initial)^2 + 2(alpha)(delta theta) to solve the problem, along with the equation I = (1/2)mr^2. However, when plugging in their own numbers, they did not get the correct answer. It is determined that the issue may be due to a negative value for alpha, resulting in
  • #1
med9546
3
0
A wheel on a game show is given an initial angular speed of 1.3 rad/s. It comes to rest after rotating through 3/4 of a turn. Find the average torque exerted on the wheel given that it is a disk of radius 0.80m and mass of 6.4 kg. I worked the matching problem in the book and got the right answer. However when I plugged in my number (the ones above) from my homework it didn't come out right. What is going wrong. I used
w^2=w(initial)^2 + 2(alpha)(delta theta)
Then I found I by using I = (1/2)mr^2
I them multiplied the two answers together to get the torque since
Torque = I(alpha)

Like I said, this worked to find the answer using the books numbers but it didn't work for my homework. (The books numbers are (W final = 0 rad/s, w initial = 1.22 rad/s, radius is 0.71m. The rest is the same. The answer is 0.25) Please help
 
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  • #2
[tex] \delta\theta=\frac{3}{4}2\pi \ \mbox{rad}=\frac{3\pi}{2} \ \mbox{rad}[/tex]

[itex]\alpha [/itex] is negative and the torque will be negative is well.

Daniel.
 
  • #3


It seems like the problem may be with the values you used for the initial and final angular speeds. In the book's problem, the wheel starts at an initial angular speed of 1.22 rad/s and comes to rest at 0 rad/s. However, in your homework problem, the initial angular speed is given as 1.3 rad/s and the final angular speed is not specified. This could be why your calculation is not giving the correct answer.

To solve the problem correctly, you'll need to find the final angular speed of the wheel. This can be done using the same formula you used, but rearranged to solve for the final angular speed:

w(final) = sqrt(w(initial)^2 + 2(alpha)(delta theta))

Plugging in the given values, we get:

w(final) = sqrt((1.3 rad/s)^2 + 2(alpha)(3/4 turn))

Since the wheel comes to rest, the final angular speed is 0 rad/s. We can solve for alpha by rearranging the formula and plugging in the values for w(final), w(initial), and delta theta:

alpha = (w(final)^2 - w(initial)^2) / (2(delta theta))

Plugging in the values, we get:

alpha = (0 rad/s)^2 - (1.3 rad/s)^2) / (2(3/4 turn))

Simplifying, we get:

alpha = -0.68 rad/s^2

Now, we can plug this value for alpha into the formula for torque:

Torque = I(alpha)

Using the given values for the wheel's mass and radius, we can calculate the moment of inertia (I) as you did:

I = (1/2)(6.4 kg)(0.80 m)^2 = 2.048 kgm^2

Plugging in the values for I and alpha, we get:

Torque = (2.048 kgm^2)(-0.68 rad/s^2) = -1.393 Nm

This answer may seem strange since torque is typically a positive value, but keep in mind that negative torque just indicates that the rotation is in the opposite direction of our chosen axis. To get the correct answer, you can simply take the absolute value of the torque:

|Torque| = 1.393 Nm

So, the average torque exerted on the wheel is 1.393 Nm
 

1. How does the wheel on a game show force problem work?

The wheel on a game show force problem is typically a large spinning wheel divided into sections with different values or outcomes. Contestants spin the wheel and the section the wheel stops on determines their prize or the action they must take.

2. What is the purpose of the wheel on a game show force problem?

The purpose of the wheel on a game show force problem is to add an element of chance or randomness to the game. It also allows for a variety of outcomes and prizes, making the game more exciting for both the contestants and the audience.

3. How is the force of the spin on the wheel determined?

The force of the spin on the wheel is typically determined by the strength and speed of the contestant's spin. Some game shows may also have a predetermined force set by the producers to ensure fairness among all contestants.

4. How is the winner determined in a game show with a wheel force problem?

The winner is determined by the outcome of the wheel spin. Depending on the game show, the winner may be the contestant who lands on the highest value or the one who fulfills a specific requirement, such as landing on a certain section of the wheel.

5. Are there any strategies for winning a game show with a wheel force problem?

Since the outcome of the wheel spin is based on chance, there is no guaranteed strategy for winning. However, some contestants may try to control the force of their spin or aim for a specific section of the wheel that they believe will increase their chances of winning.

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