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# 7.5: The Rotational Analogs of Force, Momentum, Mass and Impulse

• Contributed by Wendell Potter and David Webb et al.
• Physics at UC Davis

# The Torque Construct: The Rotational Analog to Force

Consider a force F exerted tangentially on the rim of a wheel or disk. The rim is at a distance r from the axis of rotation. We can formally define torque, represented by the Greek letter $$\tau$$ , in terms of the force F and the distance r:

$\tau = r_{\perp}F = rF_{tangential}$

where r is sometimes referred to as the moment arm of this applied force—the further away from the axis a particular force is applied, the more torque is exerted, producing more change in rotational motion. Torque can be thought of as the “turning effectiveness of a force” or “rotational force.”

Figure 7.5.1

What happens if the applied force is not purely tangential, as in the second circle in figure 7.5.1? This force can be broken down into its tangential and radial components, $$F_{tangential}$$ and $$F_{radial}$$ . Note that the radial component $$F_{radial}$$ of this force has no effect on the rotational motion of this disk! So, for any general force exerted a distance r from a rotation axis, it is only the tangential component of this force $$(F_{tangential})$$ that will affect rotational motion. The tangential component of the force can always be found with the appropriate trig function. If $$\theta$$ is the angle between the applied force, F, and r, the tangential component is $$F \sin\theta$$ .

Figure 7.5.2

Torque, along with other angular variables, has vector properties. If we imagine the torque causing the object to rotate about an axis perpendicular to the plane defined by the force and the moment arm, $$r$$ , we can use the same right-hand-rule introduced for finding the direction of $$\theta$$ and $$\omega$$ to find the direction of the torque $$\tau$$ . If you curl the fingers of your right hand in the direction of rotation that the torque would cause, then your thumb points in the direction of the torque.

# The Angular Momentum Construct:The rotational analog to momentum

It is useful to consider the angular momentum of both a point object as well as the angular momentum of extended objects. In either case, we need to be clear about the axis (or point) about which we are calculating the angular momentum

A particle with momentum p, located at position r from some point in space has angular momentum L about that point with a magnitude given by

$L=r_{\perp}p=rp_{tangential}$

Note that the angular momentum is related to the linear momentum the same way as torque is related to force. Both L and $$\tau$$ depend on the choice of the point in space to which they are referenced. Like torque, angular momentum is a vector. Its direction is perpendicular to both r and p and is given by the RHR.

Figure 7.5.3

If a system has many parts, its total angular momentum is the vector sum of the angular momenta of all the parts:

$L = L_1 + L_2 + L_3 \dots = \sum L_i$

A rigid object with rotational inertia I about some particular axis has an angular momentum about the same axis given by

$L = I\omega$

The direction of $$L$$ is parallel to the direction of $$\omega$$. These directions are shown in the figure 7.5.4.

Figure 7.5.4

# Rotational Inertia: The Rotational Analog to Mass

Recall that for translational motion an object with a large amount of inertia has a greater momentum than an object with a small amount of inertia, both moving at the same speed. Mass, m, is the measure of inertia in translational motion. The rotational motion analogy to inertia is rotational inertia (or rotational mass), or in very technical language, moment of inertia. With a given net torque, $$\Sigma\tau$$, different objects will experience different rotational accelerations.

The rotational inertia of an object does not depend solely on the amount of mass in the object, but on how this mass is distributed about the axis of rotation.

For the simplest case of a point mass m moving in a circle of radius r, its rotational inertia is given by:

$I = mr^2$

This definition allows us to calculate the rotational inertia of any object, provided we know the position r of every portion of its mass as measured perpendicularly with respect to the axis of rotation:

$I = m_1r_1^2 +m_2r_2^2 +m_3r_3^2 + \dots = \sum m_ir_i^2$

This looks a lot like calculus (which it is in the limit of infinitesimally small mass increments.) The table below gives the rotational inertia of several simple geometric shapes, as calculated in the limit of infinitesimal increments of mass using this equation.

 Object Rotational Inertia Illustration Point mass m moving in radius r I=mr2 Thin ring of mass m, radius r rotating about center I=mr2 Thin rod of mass m, length L rotating about one end perpendicular to the rod I=$$\frac{1}{3}$$mL2 Thin rod of mass m, length L rotating about one e perpendicular to the rod I=$$\frac{1}{12}$$mL2 Disk of mass s, radius r, rotating about an axis perpendicular to disk though the center I=$$\frac{1}{2}$$mr2 Sphere of mass m, radius r, rotating about an axis thorough the center I=$$\frac{2}{5}$$mr2 Thin hollow spherical shell of mass m, radius r, about an axis through the center I=$$\frac{2}{3}$$mr2

As seen from the formulas in the table, objects with the same mass can have very different rotational inertias, depending on how the mass is distributed with respect to the axis of rotation.

Also, it is possible for an object to change its rotational inertia (e.g., a gymnast tucking in or extending arms and legs), which can lead to dramatic results as net torques are applied. The rotational inertia of a composite object is the sum of the rotational inertias of each component, all calculated about the same axis.

$I_{total} = I_1 + I_2 + I_3 + \dots$

So for a ring and a disk stacked upon each other and rotating about the symmetry axis of both, the rotational inertia is :

$I_{total} = I_{ring} + I_{disk}$

The SI units of rotational inertia are $$kg \cdot{m^2}$$

# Angular Impulse: The Rotational Analog to Impulse

The angular analog to impulse, is angular impulse:

$AngJ_{ext} = \int \tau_{ext}(t)dt$

or, if the torque is constant with time, or we define an average torque $$\tau_{avg}$$

$AngJ_{ext} = \tau_{avg}\Delta t$

# A Statement of Angular Momentum Conservation:

$AngJ_{ext} = \int\tau_{ext}(t)dt = \Delta L_{system}$

or

$AngJ_{ext} = \tau_{ext}\Delta t= \Delta L_{system}$

Note

If the net external angular impulse acting on a system is zero, then there is no change in the total angular momentum of that system; otherwise, the change in angular momentum is equal to the net external angular impulse.