# 12.1. Rolling Motion

Figure 12.1. Rotational Motion of Wheel

A wheel rolling over a surface has both a linear and a rotational
velocity. Suppose the angular velocity of the wheel is [omega]. The
corresponding linear velocity of any point on the rim of the wheel is given
by_{ }

_{cm}(see Figure 12.2). We conclude that the top of the wheel moves twice as fast as the center and the bottom of the wheel does not move at all.

Figure 12.2. Motion of wheel is sum of rotational and translational
motion.

An alternative way of looking at the motion of a wheel is by regarding it as a
pure rotation (with the same angular velocity [omega]) about an instantaneous
stationary axis through the bottom of the wheel (point P, Figure 12.3).
Figure 12.3. Motion of wheel around axis through P.

## 12.2. Kinetic Energy

**The kinetic energy of the wheel shown in Figure 12.3 can be calculated easily using the formulas derived in Chapter 11**

_{ }

_{P}is the rotational inertia around the axis through P, and [omega] is the rotational velocity of the wheel. The rotational inertia around an axis through P, I

_{P}, is related to the rotational inertia around an axis through the center of mass, I

_{cm}

_{ }

_{ }

**Example Problem 12-1**

Figure 12.4 shows a disk with mass M and rotational inertia I on an inclined plane. The mass is released from a height h. What is its final velocity at the bottom of the plane ?

The disk is released from rest. Its total mechanical energy at that point is equal to its potential energy

_{ }

_{ }

_{ }

_{ }

Figure 12.4. Mass on inclined plane.

Conservation of mechanical energy implies that E_{i}= E

_{f}, or

_{ }

_{ }

_{ }

_{ }

_{ }

**the disk with the smallest moment of inertia has the largest final velocity**.

Figure 12.5. Problem 13P.

**Problem 15P**

A small solid marble of mass m and radius r rolls without slipping along a loop-the-loop track shown in Figure 12.5, having been released from rest somewhere along the straight section of the track. From what minimum height above the bottom of the track must the marble be released in order not to leave the track at the top of the loop.

The marble will not leave the track at the top of the loop if the centripetal force exceeds the gravitational force at that point:

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

**Example Problem 12-2: the yo-yo**

Figure 12.6. The yo-yo.

Figure 12.6 shows a schematic drawing of a yo-yo. What is its linear
acceleration ?There are two forces acting on the yo-yo: an upward force equal to the tension in the cord, and the gravitational force. The acceleration of the system depends on these two forces:

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

## 12.3. Torque

Figure 12.7. Motion of particle P in the x-y plane.

A particle with mass m moves in the x-y plane (see Figure 12.7). A
single force F acts on the particle and the angle between the force and the
position vector is [phi]. Per definition, the torque exerted by this force on
the mass, with respect to the origin of our coordinate system, is given by_{ }

_{ }

_{[invtee]}is called the arm of the force F with respect to the origin. According to the definition of the vector product, the vector [tau] lies parallel to the z-axis, and its direction (either up or down) can be determined using the right-hand rule. Torque defined in this way has meaning only with respect to a specified origin. The direction of the torque is always at right angles to the plane formed by the vectors r and F. The torque is zero if r = 0 m, F = 0 N or r is parallel or anti-parallel to F.

## 12.4. Angular Momentum

**The angular momentum L of particle P in Figure 12.7, with respect to the origin, is defined as**

_{ }

_{ }

A particle can have angular momentum even if it does not move in a circle. For example, Figure 12.8 shows the location and the direction of the momentum of particle P. The angular momentum of particle P, with respect to the origin, is given by

Figure 12.8. Angular momentum of particle P.

_{ }

_{ }

_{ }

**if the net torque acting on the particle is zero, its angular momentum will be constant**.

**Example Problem 12-3**

Figure 12.9 shows object P in free fall. The object starts from rest at the position indicated in Figure 12.9. What is its angular momentum, with respect to the origin, as function of time ?

The velocity of object P, as function of time, is given by

_{ }

_{ }

_{ }

Figure 12.9. Free fall and angular momentum

Figure 12.10. Action - reaction pair.

If we look at a system of particles, the total angular momentum L of the
system is the vector sum of the angular momenta of each of the individual
particles:_{ }

_{ }

_{AB}on A, than A will exert a force F

_{BA}on B. F

_{AB}and F

_{BA}are related as follows

_{ }

_{ }

_{ }

_{ }

We conclude that

_{ }

## 12.5. Angular Momentum of Rotating Rigid Bodies

**Suppose we are dealing with a rigid body rotating around the z-axis. The linear momentum of each mass element is parallel to the x-y plane, and perpendicular to the position vector. The magnitude of the angular momentum of this mass element is**

_{ }

_{ }

_{ }

_{ }

**Only if the rotation axis is a symmetry axis of the rigid body will the total angular momentum vector coincide with the rotation axis.**

## 12.6. Conservation of Angular Momentum

**If no external forces act on a system of particles or if the external torque is equal to zero, the total angular momentum of the system is conserved. The angular momentum remains constant, no matter what changes take place within the system.**

**Problem 54E**

The rotational inertia of a collapsing spinning star changes to one-third of its initial value. What is the ratio of the new rotational kinetic energy to the initial rotational kinetic energy ?

The final rotational inertia I

_{f}is related to the initial rotational inertia I

_{i}as follows

_{ }

_{ }

_{ }

_{ }

Figure 12.11. Problem 61P.

**Problem 61P**

A cockroach with mass m runs counterclockwise around the rim of a lazy Susan (a circular dish mounted on a vertical axle) of radius R and rotational inertia I with frictionless bearings. The cockroach's speed (with respect to the earth) is v, whereas the lazy Susan turns clockwise with angular speed [omega]

_{0}. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops ? (b) Is mechanical energy conserved ?

Assume that the lazy Susan is located in the x-y plane (see Figure 12.11). The linear momentum of the cockroach is m

^{.}v. The angular momentum of the cockroach, with respect to the origin, is given by

_{ }

_{ }

_{0}is less than zero. The total angular momentum of the system is given by

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

## 12.7. The Precessing Top

Figure 12.12. The precessing top.

A top, set spinning, will rotate slowly about the vertical axis.
This motion is called precession. For any point on the rotation axis of the
top, the position vector is parallel to the angular momentum vector.The weight of the top exerts an external torque about the origin (the coordinate system is defined such that the origin coincides with the contact point of the top on the floor, see Figure 12.12). The magnitude of this torque is

_{ }

_{ }

_{ }

_{ }

^{2}and the precession is independent of the angle [theta].

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