# Rotational Acceleration

1. Jan 27, 2005

### BrianD

Hello,
I'm hoping someone can help me understand the physics involved in a rotating object.

Imagine 2 bicycle tires of different sizes with string wrapped around the rims.

How would I describe or calculate which tire requires less force to start spinning, or the change in force necessary to accomodate different diameter tires?

I would also like to understand how to describe what a change in a tires mass distribution does to affect the force needed for startup. ie, what happens when weight is added near the center of the tire as opposed to around the rim?

My appologies for a very basic question. Thank you for your help!

Best,

Brian

2. Jan 27, 2005

### vinter

Where is the bicycle wheel in your question? on a frictionless surface, in a gravity free place, the wheel is standing straight, lying on the floor????

3. Jan 28, 2005

### Justin Lazear

4. Jan 28, 2005

### FredGarvin

Like Justin said, it's all basically in the mass moment of inertia and the rotational form of Newton's second law.

5. Jan 28, 2005

### da_willem

Hi BrainD, welcome to Physics Forums!

I take you're interested only in the force needed to bring the wheel into motion, ignoring friction forces acceleration of the entire bike etc. I don't know how much physics or mathematics you know so I advise you to indeed read up on moment of inertia. It is a concept related to mass, but it takes into account the mass distriution. You already noticed the effect that mass close to the axis of rotation is easier to give a certain cycles per second than mass further away from the axis. As you might know accelerating mass requires a force F wich is proportional to both the acceleration a and the mass m by Newtons second law:

$$F=ma$$

For circular motion of a rigid body a similar formula exists:

$$T=I\alpha$$

Now the force is replaced by a 'torque' T, mass by the 'moment of inertia' I and acceleration by 'angular acceleration' $\alpha[/tex]. If you really want to find the answer to your question I advise you to read up on all three of them (eg try 'googling' them). 6. Jan 29, 2005 ### BrianD A second component First off, thank you all very much for your help. If I could trouble you help with one more aspect of this problem, as moment of inertia alone does not acurately explain what I've observed. I'll have to call this second component "leverage", although i'm sure that's not right, and am refering to the fact that as the radius of the wheel get's larger force applied to it's outside makes it easier to set in motion. In the bicycle wheel example, the wheel is vertical (and here on earth, but I'm hoping friction is constant so I only have to consider 2 forces), and is only the tire's rim (no bicycle, and no rubber on the rim). First, I'll wrap a rope around the rim 10 times, and pull the rope with a fixed amount of force to set the wheel spinning. The next time, I wrap the rope around the rim 100 times (It's a very deep rim!), and pull with the same amount of force. In this case, I've increased the mass at the outer edge of the wheel to increase it's moment of inertia, but I've also increased the radius of the wheel (by wrapping more rope around it), making it easier to set into motion. I guess what the question boils down to is, what is the proper name for this "leverage", and how can I look at and compare the effects of both in getting this wheel to start spinning? Thank you very much for your help! Best, Brian 7. Jan 29, 2005 ### da_willem Lets consider an axially symmetric mass distribution. This is aproximately true in your case if you assume the wheel is much more massive than the rope one time around the wheel. This simplifies things considerably as it makes the moment of inertia much more simple and also we dont need to consider (much) the vector character of the involved quantities. Torque: In this case the torque needed is simply force times the 'arm', the point from the center of mass to the point you pull. [itex]T=rF[/tex] Moment of inertia The definition is[itex] I=\int r^2 dm$ If you think of all th mass distributed a distance R from th center of the wheel this is just $R^2 M$ But ofcourse it's not that simple and I will increase by adding mass, and also if you distribute the mass further from the center

Now as I stated earlier the equation of motion (with \lpha the angular acceleration) is:

$$T=I\alpha$$

So the force needed to set the wheel (radius R) in motion is:

$$F=\frac{T}{R}=\frac{I \alpha}{R}$$

As you increase R the force needed to set the wheel in motion decreases, but at the same time I increases making it hardre to increase the angular acceleration. Thus describing the effect you mention...