# Rotational mechanics

1. Sep 29, 2009

### shohin

hi,
a doubt regarding rolling motion of a ball down the incline has been giving me sleepless nights for a couple of days now, here is it-

if a perfect sphere is placed on an inclined plane with angle of inclination x (say), its translation of the CM, and rotation can be treated seperately, now if the frictional force acting on the bottommost point equals the gravity component down the incline, then the centre of mass(CM) should not accelerate as they result into 0, but however the torque would exist causing the ball to rotate(gravity will not create torque offcourse), this means that the body wud not move down but rotate at the fixed position with slipping,
However this does not happen as the ball moves down with rolling
WHY????
tnk u

2. Sep 29, 2009

### Staff: Mentor

If you figure out (using Newton's laws) the friction required to prevent slipping, you'll see that it's always less than the component of gravity down the incline.

3. Sep 29, 2009

### shohin

thats good, but i wwnted to knw what would happen if i placed a perfect sphere of mass
'm', on an incline of inclination x, such that the frictional force equals mgsinx (component of gravity),

4. Sep 29, 2009

### Staff: Mentor

My point is that you can't do that. The friction force is whatever it needs to be to prevent slipping, which will always be less than mgsinx. You may choose the materials, but you don't choose the friction force--it's set by the physics and geometry of the sphere and incline.

5. Sep 29, 2009

### shohin

i got this frm another source, it wrong then acoording to u????

"To explain, let me first give you the example of a block resting on a table with friction. The block does not move till the force is more than the maximum frictional force even though we may have applied force passing through its center of mass which results into torque. That is because, the force applied passing throgh the center of mass will have the tendency to topple the block till it is less than the frictional force. This will cause the block to lose contact with the table surface except at the front edge. This will cause the block to get slightly lifted up towards the ;point of application of force, but the torque due to applied force and due to gravity about the contact point will balance. So it neither moves nor rolls over.

Now coming to the case of a sphere, suppose the sphere is kept at one end of the incline and the angle of incline is slowly raised. Upto some angle, the frictional force, μmgcosθ is more than the component of weight, mgsinθ. Hence no translation of center of mass occurs. As it rolls, the line joining the center and the contact point with the surface rotates and does not move parallel to itself. In other words, in one rotation distance covered by the contact point = circumference which is a purely rotational motion. This is the situation till μmgcosθ equals mgsinθ. Thereafter mgsinθ is more than μmgcosθ. Now onwards, both rolling and slipping should occur. It means that with one rotation, distance covered is more than the length of circumference. The distance covered in one rotation minus the length of circumference is the distance covered due to translatory motion and that divided by time the translatory velocity. Total velocity of the center of mass minus this translatory velocity when multiplied by radius will give the angular velocity in this case.

Summarizing, the sphere has purely rotatory motion upto some angle θ given by μmgcosθ = mgsinθ and both translatory and rotatory motion thereafter. It may appear to you that till that angle also center of mass is moving. So how can we say that there is no translatory motion. To understand that, it is necessary to understand the definition of translatory motion.

A body is said to have a purely translatory motion when a line joining any two of its points moves parallel to itself. This does not happen in the case of a sphere rolling down upto some angle. Beyond that angle, its velocity is such that vR > ω and (vR - ω)/R is the velocity due to translatory motion. This is the velocity with which the point of contact of the sphere slips when angle of inclination is more than certain angle.

Another way of explaining is that the contact point is instantaneously stationary till rotation without slipping. When rotation occurs combined with slipping, the contact point has an instantaneous linear velocity which describes its translatory motion."

6. Sep 29, 2009

### Staff: Mentor

What source?

That's completely wrong. The friction force does not equal μmgcosθ. (Your source is confusing the maximum static friction between two surfaces, which is μN, with the actual friction force, which in this case is less.)

7. Sep 29, 2009

### shohin

why is it less? why cant it be equal?
static friction should always equal the opposite force

8. Sep 29, 2009

### Staff: Mentor

In general, static friction can range from 0 to the maximum value of μN. The value it assumes in any given situation depends on the specifics of that situation.
I'm not sure what you mean. Certainly the friction force that the surface exerts on the sphere will be equal and opposite to the force that the sphere exerts on the surface.

Note that friction acts to prevent slipping between surfaces. In the case of a sphere rolling down an incline, the required friction force to prevent slipping is less than the component of gravity down the incline.

9. Sep 29, 2009

### shohin

"the required friction force to prevent slipping is less than the component of gravity down the incline."

can u prove this???

10. Sep 29, 2009

### Staff: Mentor

Sure. Just apply Newton's 2nd law to both rotation and translation. The forces acting on the sphere are: gravity (mg), friction (F), and the normal force (N).

For translation:
mgsinθ - F = ma

For rotation:
Fr = Iα
Fr = 2/5 mr^2 (a/r)
F = 2/5 ma

Combine and solve for F:
mgsinθ - 2/5 ma = ma
ma = 5/7 mgsinθ
Thus: F = 2/7 mgsinθ

11. Sep 29, 2009

### shohin

tnxxxxxxxxxxxxxxxxxxxxx a lotttttttttt for your time
ur a life saver
i think i finally got it