What is the rotational inertia of a model rocket attached to a rigid rod?

[rage_pandora]

Hi,

This is my first time posting on this site and I was wondering if someone could help me with this question:

A 200 g model rocket shown in Figure Ex13.24 generates 4.0 N of thrust. It spins in a horizontal circle at the end of a 100 g rigid rod. What is its angular acceleration? (I have attached the figure below).
http://www.pandorainc.com/physicspics/DSCF0061.JPG

Thanks.

Rage

 

[Doc Al]

You'll need to apply Newton's 2nd law for rotation. What's the torque exerted by the rocket? What's the rotational inertia of the system?

 

[OlderDan]

Welcome to the forum. What we do here is help people who have shown us their effort to solve a problem, not just solve problems for them. Show us what you think needs to be considered/done to solve this problem, and then we will help you with it.

 

[rage_pandora]

Ok so basically I know that I'm supposed to use the formula: net torque = I*a. I also know that the torque will be r*F*sin(45). The problem I am having is figuring out whether I use the whole length(0.6m) for the radius, or the center of mass of the system? The other thing I am stuck on is calculating the moment of inertia. I made "I" equal to the total mass of the system (0.3kg) times the distance to the center of mass squared. I am not sure if this is right or do I have to, again , separate each object into its own radius (m1*r1^2 + m2*r2^2). Thanks.

 

[Doc Al]

Ok so basically I know that I'm supposed to use the formula: net torque = I*a. I also know that the torque will be r*F*sin(45). The problem I am having is figuring out whether I use the whole length(0.6m) for the radius, or the center of mass of the system?

That radius vector represents the distance from the center to the point of application of the force, so what do you think?

The other thing I am stuck on is calculating the moment of inertia. I made "I" equal to the total mass of the system (0.3kg) times the distance to the center of mass squared. I am not sure if this is right or do I have to, again , separate each object into its own radius (m1*r1^2 + m2*r2^2).

Rotational inertia depends on the distribution of the mass, not just the location of the center of mass. After all, a ball and a stick could have the same mass and center of mass, but would they have the same rotational inertia?

Yes, you must add the rotational inertia of each object together to find the total rotational inertia. What's the rotational inertial of a rod about one end?

 

[rage_pandora]

Ok so to find the net torque I multiplied the whole radius (0.6m) by the force (4N) and sin (45) which gave me a final value of 1.697 Nm. For the rotational inertia I added the rotational inertia of a rod about one end (1/3)(M)L^2 and the rotational inertia of the rocket mr^2 which gave me a final value of 0.084 kg m^2. Dividing these two values gave me a rotational acceleration of 20.2 rad/s^2 which seems about right. I still don't understand why though we are taking them as separate objects when finding rotational inertia because I would think that since they are attached you could combine the two and take the rotational inertia of the center of mass of the whole system?

 

[Doc Al]

I still don't understand why though we are taking them as separate objects when finding rotational inertia because I would think that since they are attached you could combine the two and take the rotational inertia of the center of mass of the whole system?

You still seem to think that the rotational inertia of an object can be found by treating its mass as if concentrated at the center of mass. In general that's not true. That doesn't even work for the stick by itself, never mind the combination of stick plus rocket. (True, we are ignoring the length of the rocket, since it's not given. But if it's small enough compared to the distance to the center, that should be fine.)

As I said earlier, the rotational inertia depends on how the mass is distributed, not just on the location of the center of mass. After all, a stick rotating about its center has rotational inertia, despite the fact that its center of mass is right on the axis.