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Physics · Ch 6 — System of Particles and Rotational Motion

Kinematics and Dynamics of Rotational Motion About a Fixed Axis

6.14

Kinematics and Dynamics of Rotational Motion About a Fixed Axis

For a rigid body rotating about a FIXED axis with a CONSTANT angular acceleration α\alpha, the angular

displacement θ\theta, angular velocity ω\omega and angular acceleration α\alpha obey three equations

that are exact rotational counterparts of the familiar straight-line kinematic equations, obtained simply

by replacing every linear quantity with its angular counterpart (s→θs\to\theta, v→ωv\to\omega, a→αa\to\alpha;

see the full correspondence in Section 5.15):

ω=ω0+αt\omega = \omega_0 + \alpha t

θ=ω0t+12αt2\theta = \omega_0 t + \frac12 \alpha t^2

ω2=ω02+2αθ\omega^2 = \omega_0^2 + 2\alpha\theta

where ω0\omega_0 is the angular velocity at time t=0t=0. These three equations are used in exactly the same

way, and under exactly the same restriction (constant angular acceleration only), as their straight-line

counterparts from earlier chapters.

Rotational dynamics answers the separate question of WHAT PRODUCES a given angular acceleration.

Exactly as Newton's second law F=maF = ma connects a net force to the linear acceleration it produces in a

body of mass mm, its rotational counterpart connects a net external torque to the angular acceleration it

produces in a rigid body of moment of inertia II, about the same fixed axis:

τ=Iα\tau = I\alpha

This equation, together with the moment of inertia values from Section 5.11 and the kinematic equations

above, is what makes it possible to solve a complete rotational problem end to end -- for example, finding

how long a constant applied torque takes to spin a flywheel up from rest to some target angular velocity,

or how quickly a constant frictional torque brings a freely spinning wheel to a stop, exactly as worked out

in this chapter's numerical problems. The corresponding rotational kinetic energy of a body spinning at …