Need help/varify on a torque/moment of inertia problem

[iamtrojan3]

Homework Statement

Hi, this is my first time on this forum, so don't flame me for not doing things right. I'm a junior in high school and currently taking AP physics, this problem in my hwk has been bothering me for a while. I've never received the answer to this question and i just need to know if i did this right, since this is really the first moment of inertia problem i encountered...
the problem is:
A force of 15N is applied tangentially to the edge of a 0.88kg solid disc initially at rest. The radius of the disc is 0.55m. How fast will the disc be spinning after it has gone 3.0 complete rotations? (disregard all the friction/air resistance etc.)

Homework Equations

Torque = I (alpha) from F=ma
W^2= Wo^2 + 2 (alpha)(theta)
Torque= Lever arm x Force applied

The Attempt at a Solution

The moment of Inertia of a solid disc is I=1/2mr^2 >> I=1/2(0.88)(0.55^2) >> I=0.1331
Torque= Lever arm x Force applied Torque= 15N x 0.55m = 8.25Nm
Torque = alpha x I >>> 8.25= alpha x 0.1331 >>> alpha = 62 (this doesn't look right )
Then just use Kenematic Equations
since 3 rotations is 6pi
W^2=Wo^2x2(alpha)(Theta)
W^2=0(62)(6pi)>>> W= 48.35radians/sec (this doesn't look right either )

Feel free to correct me on watever...i admitt, i suck/hate these problems.
Thanks

 

[Mindscrape]

Looks okay to me, but it is always easy to follow someone's work and make the same simple mistake they did.

 

[m2003]

Hi there,

First of all, it's great that you are seeking help and verifying your work. That shows a strong dedication to understanding the material and improving your skills. Let's take a look at your solution and see if we can make any corrections or improvements.

Your moment of inertia calculation looks correct. It's important to note that the moment of inertia of a solid disc is actually 1/2mr^2, where m is the mass and r is the radius. This is because the disc has a uniform mass distribution, so the moment of inertia is calculated using the average radius (0.55m in this case).

Next, let's look at your calculation of torque. You correctly used the formula torque = lever arm x force applied. However, the lever arm in this case is not the radius of the disc, but rather the distance from the pivot point (center of the disc) to the point where the force is applied. In this case, the force is applied tangentially to the edge of the disc, so the lever arm is equal to the radius of the disc (0.55m). Therefore, the torque should be 15N x 0.55m = 8.25Nm.

Now, let's move on to the calculation of angular acceleration (alpha). You correctly used the formula torque = alpha x moment of inertia. However, your units are not consistent. Torque is measured in Nm, while moment of inertia is measured in kgm^2. In order to get the units to match, you need to convert the mass from kg to N (multiply by 9.8m/s^2, the acceleration due to gravity). This gives you a torque of 8.25Nm, and a moment of inertia of 8.63kgm^2. Now, you can solve for alpha: 8.25Nm = alpha x 8.63kgm^2, which gives you an angular acceleration of 0.96 rad/s^2.

Finally, let's look at your use of the kinematic equation. You correctly used the formula w^2 = wo^2 + 2(alpha)(theta). However, you made a small mistake in your calculation of theta. Each rotation is equal to 2pi radians, so 3 rotations is equal to 6pi radians. Plugging in the values, we get w^2 = 0^2 + 2(0.96