How Do You Calculate the Rotational Inertia of a Wheel?

[nicolec08]

Hi everyone, I'm having a little trouble trying to answer this problem. Here it is:

A force of 22.04 N is applied tangentially to a wheel of radius 0.340 m and gives rise to an angular acceleration of 1.20 rad/s^2. Calculate the rotational inertia of the wheel.

Okay so i attempted the problem, here's what I got.

F=22.04
r= 0.340
m=?
$$\alpha$$=1.20 rad/s^2
$$\tau$$=Fxr = 7.49

And now I don't know where to finish. All I know is that you need torque to get the moment of inertia, and I need a mass to find the moment of inertia...Help me please!

 

[Doc Al]

And now I don't know where to finish. All I know is that you need torque to get the moment of inertia, and I need a mass to find the moment of inertia...Help me please!

If you have the torque and the angular acceleration, you don't need the mass to find the moment of inertia. Hint: How would you write Newton's 2nd law for rotational motion?

 

[nicolec08]

torque = Tdsin (theta)?

 

[Doc Al]

torque = Tdsin (theta)?

No, that's just the definition of torque.

Newton's 2nd law for translational motion is: F = ma

How would you write the equivalent law for rotational motion? Hint: What would F, m, and a be replaced with?

 

[rwisz]

Hi there, great job on attempting the problem! To find the rotational inertia of the wheel, also known as its moment of inertia, we need to use the formula I = \frac{mr^2}{2}, where m is the mass of the wheel and r is the radius. We can rearrange this formula to solve for m: m = \frac{2I}{r^2}.

Now, we can use the torque equation \tau = I\alpha to find the moment of inertia I. Plugging in the values we know, we get:

7.49 = I(1.20)

Solving for I, we get I = 6.24 kg·m^2.

Finally, we can plug this value of I into the formula we found earlier to solve for the mass of the wheel: m = \frac{2(6.24)}{0.340^2} = 110.59 kg.

So, the rotational inertia of the wheel is 6.24 kg·m^2 and the mass of the wheel is 110.59 kg. I hope this helps! Just remember to always use the appropriate formulas and units when solving physics problems. Keep up the good work!