Miss

Topics: Angular momentum, Classical mechanics, Mass Pages: 19 (2308 words) Published: November 22, 2013
Chapter 11
Rolling, Torque, and Angular Momentum
In this chapter we will cover the following topics:
-Rolling of circular objects and its relationship with friction -Redefinition of torque as a vector to describe
rotational problems that are more complicated than the rotation of a rigid body about a fixed axis
-Angular momentum of single particles and
systems of particles
-Newton’s second law for rotational motion
-Conservation of angular momentum
-Applications of the conservation of angular momentum
(11-1)

t1 = 0

t2 = t

Rolling as Translation and Rotation Combined
Consider an object with circular cross section that rolls
along a surface without slipping. This motion, though
common, is complicated. We can simplify its study by
treating it as a combination of translation of the center of mass and rotation of the object about the center of mass.

Consider the two snapshots of a rolling bicycle wheel shown in the figure. An observer stationary with the ground will see the center of mass O of the wheel move forward with a speed vcom . The point P at which the wheel makes contact with the road also moves with the same speed. During the time interval t between ds

the two snapshots both O and P cover a distance s, vcom =
(eq. 1). During t
dt
the bicycle rider sees the wheel rotate by an angle θ about O so that ds

s = Rθ →
=R
=ω (eq. 2). If we combine equation 1 with equation 2
dt
dt
we get the condition for rolling without slipping: vcom = Rω (11-2)

vcom = Rω

We have seen that rolling is a combination of purely translational motion with speed vcom and a purely rotational motion about the center of mass vcom
. The velocity of each point is the vector sum
R
of the velocities of the two motions. For the translational motion the r
velocity vector is the same for every point (vcom ,see fig. b). The rotational with angular velocity ω =

velocity varies from point to point. Its magnitude is equal to ω r where r is the distance of the point from O. Its direction is tangent to the circular orbit (see fig. a). The net velocity is the vector sum of these two terms. For example, the velocity of point P is always zero. The velocity of the center of mass O is r

r
vcom (r = 0). Finally, the velocity of the top point T is equal to 2vcom . (11-3)

vT
A
vO

vA
B
vB

Rolling as Pure Rotation
Another way of looking at rolling is shown in the figure.
We consider rolling as a pure rotation about an axis
of rotation that passes through the contact point P
between the wheel and the road. The angular
v
velocity of the rotation is ω = com .
R

In order to define the velocity vector for each point we must know its magnitude as well as its direction. The direction for each point on the wheel points along the r
tangent to its circular orbit. For example, at point A the velocity vector v A is perpendicular to the dotted line that connects point A with point B. The speed of each point is given by v = ω r. Here r is the distance between a particular point and the contact point P. For example, at point T r = 2 R. Thus vT = 2 Rω = 2vcom . For point O r = R, thus vO = ω R = vcom . For point P, r = 0 thus vP = 0.

(11-4)

The Kinetic Energy of Rolling
Consider the rolling object shown in the figure.
It is easier to calculate the kinetic energy of the rolling
body by considering the motion as pure rolling
about the contact point P. The rolling object has mass M
and radius R.
1
The kinetic energy K is then given by the equation K = I Pω 2 . Here I P is the 2
rotational inertia of the rolling body about point P. We can determine I P using

1
( I com + MR 2 ) ω 2 .
2
1
1
1
1
1
2
2
2
2
2
2 2
K = I comω + Mvcom
K = ( I com + MR ) ω = I comω + MR ω
2
2
2
2
2
The expression for the kinetic energy consists of two terms. The first term corresponds to the rotation about the center of mass O with angular velocity ω. the parallel axis theorem. I P = I com + MR 2 → K =...
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