Wheel on a game show force problem

[med9546]

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

 

[dextercioby]

$$\delta\theta=\frac{3}{4}2\pi \ \mbox{rad}=\frac{3\pi}{2} \ \mbox{rad}$$

$\alpha$ is negative and the torque will be negative is well.

Daniel.

 

[thepatientmental]

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