Section 10.8: Energy in Rotational Motion

Translation-Rotation Analogues & Connections Displacement x θ Velocity v ω Acceleration a α Force (Torque) F τ Mass (moment of inertia) m I Newton’s 2nd Law ∑F = ma ∑τ = Iα Kinetic Energy (KE) (½)mv2 (½)Iω2 CONNECTIONS: v = rω, atan= rα aR = (v2/r) = ω2r , τ = rF , I = ∑(mr2)

The radial component of F Work done by force F on an object as it rotates through an infinitesimal distance ds = rdθ dW = Fds = (Fsinφ)rdθ dW = τdθ The radial component of F does no work because it is perpendicular to the displacement.

Power The rate at which work is being done in a time interval Δt is This is analogous to P = Fv for translations.

Work-Kinetic Energy Theorem The work-kinetic energy theorem in rotational language states that the net work done by external forces in rotating a symmetrical rigid object about a fixed axis equals the change in the object’s rotational kinetic energy

Ex. 10.11: Rod Again

Sect. 10.9 Rolling Objects The curve shows the path moved by a point on the rim of the object. This path is called a cycloid The line shows the path of the center of mass of the object In pure rolling motion, an object rolls without slipping In such a case, there is a simple relationship between its rotational and translational motions

Rolling Object The velocity of the center of mass is The acceleration of the

A point on the rim, P, rotates to various positions such as Q and P. At any instant, a point P on the rim is at rest relative to the surface since no slipping occurs Rolling motion is thus a combination of pure translational motion and pure rotational motion

Total Kinetic Energy K = (½)Mv2 + (½)Iω2 The total kinetic energy of a rolling object is the sum of the translational energy of its center of mass and the rotational kinetic energy about its center of mass K = (½)Mv2 + (½)Iω2 Accelerated rolling motion is possible only if friction is present between the sphere and the incline

Example:   y = 0 Sphere rolls down incline (no slipping or sliding). KE+PE conservation: (½)Mv2 + (½)Iω2 +MgH = constant, or (KE)1 +(PE)1 = (KE)2 + (PE)2 where KE has 2 parts: (KE)trans = (½)Mv2 (KE)rot = (½)Iω2 v = 0 ω = 0  v = ?   y = 0

Summary of Useful Relations

Translation-Rotation Analogues & Connections Displacement x θ Velocity v ω Acceleration a α Force (Torque) F τ Mass (moment of inertia) m I Newton’s 2nd Law ∑F = ma ∑τ = Iα Kinetic Energy (KE) (½)mv2 (½)Iω2 CONNECTIONS: v = rω, atan= rα aR = (v2/r) = ω2r , τ = rF , I = ∑(mr2)

Ex. 10.12: Energy & Atwood Machine