Physics 141 Mechanics Lecture 14 Rotation Yongli Gao

1. Physics 141 Mechanics Lecture 14 Rotation Yongli Gao • Suppose we have a reference line perpendicular to the rotation axis. The motion of the rotating body can be described by q, the angular position of the line. If r is the radius of the circle, s the length of the arc, The angular position , in radians. • The circumference of a circle of radius r is 2pr, so => 1 rad = 57.3° • The angular displacement is the change in angular position , again in radians.

2. Angular Velocity and Angular Acceleration • The angular velocityw(t) is the rate of angular displacement, , unit rad/s. • The angular accelerationa(t) is the rate of angular velocity change, , unit rad/s2. • In a uniform circular motion, w=const. a=0, period T=2p/w, tangential speed v=wr, centripetal acceleration a=w2r=v2/r. • If a = const,

3. Angular Velocity as Vector • The angular velocity w is in fact a vector described by the right hand rule: if your curl your right hand with fingers pointing at the direction of rotation, your thumb is the direction of w. • If one point at the rotational axis is the origin of the the coordinate system, the velocity v at any position r of the rotating object is the cross product v=wxr. • Note that finite angular displacement is not a vector: the sequence here matters. w v r

4. Acceleration in Rotation • The linear acceleration a at position r in rotation can have two components : the centripetal accelerationar and tangential accelerationat. a=ar+at • The centripetal acceleration ar at the position r is ar=wxv=wx(wxr) (a=w2r=v2/r) • The tangential acceleration at is the cross product of angular acceleration a and the position r at=axr • The direction of angular acceleration a is described by the right hand rule: if your curl your right hand with fingers pointing at the direction of tangential acceleration, your thumb is the direction of a.

5. Angular Momentum • In rotation, angular momentumL takes the place of linear momentum p in linear motion. If a particle is located at position r, its angular momentum with respect to the origin is given by the cross product • L=pd, L=ptr • If a particle of mass m is in a uniform circular motion of radius R and speed v, it’s angular momentum with respect to the center of the circle is L=mvR(=mwR2) • Even if a particle m is moving with a linear velocity v, one can still calculate it’s angular momentum, which depends also on where the origin is chosen. p=mv pt=psinq q pr r d=rsinq

6. Torque • The torquet of a force F acting on a particle at position r is given by the cross product, (with respect to the origin) • One can understand the definition of t by realizing the effect of the force to rotate is only from the the component perpendicular to r, t=Ftr, or the distance of the force to the origin, t=Fd. • The significance of the third element of a force, the point of action, is obvious when one has to calculate the torque. • Try to open a door with the same magnitude and direction of force, but acting on different locations. F Ft q Fr r O

7. The distance from the force to the axis, d, is also called moment arm, from whicht=Fd. From the right hand rule of cross product, if we are looking in the direction of the axis, the torque is positive if it causes counter clockwise rotation (CCW). • The torque by gravity near the surface of the Earth, where g=const. is that as if the the total weight is at CM. F1 F2 d1 d2 O F3 mg dCM

8. Angular Momentum and Torque • It can be easily shown that rate of change of angular momentum L of a particle about a rotation axis equals to the torque t of the total force F acting on the particle about the same axis • The above relation is in fact the Newton’s 2nd law in rotational motion. • The unit of torque in SI is Newton meter.

9. Demonstration: Spinning Wheel and Gravity A wheel supported at one end of the axis will fall sideways because of the gravity when not rotating. When spin at high speed, the gravity acting on the wheel will produce a torque so that the spinning wheel will process. L t (inward) r mg

10. Conservation of Angular Momentum • Analogue of the Newton’s third law of motion: the torque exerted on body A by body B is equal in magnitude, but opposite in direction, to the torque exerted on B by A. • This is not an independent law of physics, but can be derived from the Newton’s third law. • It follows that for any system, what affects the total angular momentum is the total external torque: • For an isolated system, the external torque is zero, and we get an important law of physics: the conservation of angular momentum for any isolated system.

11. Examples of Conservation of Angular Momentum • A helicopter has main blades and tail blades. • Another solution is two sets of main blades of opposing spins. • A bullet is made to spin by rifling, which makes it more stable because of conservation of angular momentum. • A good quarterback always throw his/her ball with a good spin. • It’s hard to keep a bicycle at rest steady, but much easier as you start moving.

12. Example: Conservation of Angular Momentum In the horizontal plane, a particle of mass m and velocity v hits and stuck to another one of mass M at rest and attached by a massless rod of length l to a pivotal point. What should be the final velocity V? Solution: The only external force must be from the pivotal point, with respect to which no external torque => angular momentum of the system conserved V is perpendicular to l immediately after the collision. Note the linear momentum is not conserved here. l M v q m