Welcome back to Physics 211

Welcome back to Physics 211 Today’s agenda: Rotational Dynamics Kinetic Energy Angular Momentum

Reminder • Exam 3 in class Thursday, November 17 • HW 9 due Wednesday, November 16 • Blue homework book HW - 65, 66, 72 • MPHW 6 due Friday, November 18 - assignment will be online on Friday, November 11 elastic energy, linear momentum, collisions, center of mass, equilibrium of rigid bodies, torque, rotational dynamics, angular momentum

Recap • Torque  tendency of force to cause rotation • Angular velocity, acceleration for rigid body rotating about axis

A disk is rotating at a constant rate about a vertical axis through its center. Point Q is twice as far from the center as point P. The angular velocity of Q is • twice as big as P • the same as P • half as big as P • none of the above

A disk is rotating at a constant rate about a vertical axis through its center. Point Q is twice as far from the center as point P. The linear velocity of Q is • twice as big as P • the same as P • half as big as P • none of the above

Rotating disk demo

Computing torque F |t|= |F|d = |F||r|sinq = (|F| sinq)|r q r d component force at 90o to position vector times distance O

Balancing stick demo

Rotational Motion w * Particle i: ri |vi| = ri w at 90o to ri Fi pivot mi * Newton’s 2nd law: miDvi/Dt=FiT component at 900 to ri * Substitute for vi and multiply by ri : miri2Dw/Dt = FiT ri = ti * Finally, sum over all masses: Dw/DtSmiri2 = Sti = tnet

Discussion Dw/Dt (Smiri2)= tnet a - angular acceleration Moment of inertia, I I a = tnet compare this with Newton’s 2nd law Ma = F

Moment of Inertia * I must be defined with respect to a particular axis

Demo • Spinning a weighted bar – moments of inertia

Dm r Moment of Inertia of Continuous Body Dma0

Tabulated Results for Moments of Inertia of some rigid, uniform objects (from p.342 of University Physics, Young & Freedman)

D D Parallel-Axis Theorem CM *Smallest I will always be along axis passing through CM

Practical Comments on Calculation of Moment of Inertia for Complex Object • To find I for a complex object, split it into simple geometrical shapes that can be found in Table 9.2 • Use Table 9.2 to get ICMfor each part about the axis parallel to the axis of rotation and going through the center-of-mass • If needed use parallel-axis theorem to get I for each part about the axis of rotation • Add up moments of inertia of all parts

Example a axis Find moment of inertia of the object, consisting of the square plate with mass ms=2kg and width a=0.5m and of the disk with mass md=1.5kg and diameter equal to a, about the axis of rotation perpendicular to the plate and going through the center of the side of the square plate as indicated. a a

Conditions for equilibrium of an extended object For an extended object that remains at rest and does not rotate: • The net force on the object has to be zero. • The net torque on the object has to be zero.

Beam problem CM of beam N r r rm x m M? Mb = 2m Vertical equilibrium? Rotational equilibrium?

Suppose M replaced by M/2 ? vertical equilibrium? rotational dynamics? net torque? which way rotates? initial angular acceleration?

Moment of Inertia ? I = Smiri2: replace plank by point mass situated at CM * depends on pivot position! I = * Hencea=t=

Constant angular acceleration * Assume a is constant: • Dw/Dt = a i.e (wF - wI)/t = a • wF = wI + at * Now (wF + wI)/2 = wav if constant a * Then with qF - qI = wavt : qF= qI+ wIt +1/2 a t2

Problem – slowing a DVD wI = 27.5 rad/s, a= -10.0 rad/s2 how many revolutions per second ? linear speed of point on rim ? angular velocity at t = 0.3s ? when will it stop ? 10 cm

Rotational Kinetic Energy K = Si(1/2mivi2 = (1/2)w2Simiri2 Hence K= (1/2)Iw2 Energy rigid body possesses by virtue of rotation

Simple problem Cable wrapped around cylinder. Pull off with constant force F. Suppose unwind a distance d of cable 2R F what is final angular speed of cylinder ?

cylinder+cable problemenergy method • Use work-KE theorem • work W = ? • Moment of inertia of cylinder ? • from table:  =

cylinder+cable problemconst acceleration method extended free body diagram N F * no torque due to N or FW * why direction of N ? * torque due to t= FR * hence a= FR/[(1/2)MR2] = 2F/(MR)  w FW radius R

Angular Momentum • can define rotational analog of linear momentum called angular momentum • in absence of external torque it will be conserved in time • True even in situations where Newton’s laws fail ….

Definition of Angular Momentum * Back to slide on rotational dynamics: miri2Dw/Dt = ti * Rewrite, using li = miri2w Dli/Dt = ti * Summing over all particles in body: DL/Dt = text L – angular momentum = Iw

Demos • Rotating stand plus dumbbells – spin faster when arms drawn in

Rotational Motion w * Particle i: ri |vi| = ri w at 90o to ri Fi pivot mi * Newton’s 2nd law: miDvi/Dt=FiT component at 900 to ri * Substitute for vi and multiply by ri : miri2Dw/Dt = FiT ri = ti * Finally, sum over all masses: Dw/DtSmiri2 = Sti = tnet

Points to note • If text = 0, L = Iw is constant in time • conservation of angular momentum • Internal forces/torques do not contribute to external torque. • L = mvr if v is at 900 to r for single particle • L = r x p -- general result (x= vector cross product)

The angular momentum L of a particle • is independent of the specific choice of origin • is zero when its position and momentum vectors are parallel • is zero when its position and momentum vectors are perpendicular • is zero if the speed is constant

An ice skater spins about a vertical axis through her body with her arms held out. As she draws her arms in, her angular velocity • 1. increases • 2. decreases • 3. remains the same • 4. need more information

Welcome back to Physics 211