复旦大学：《大学物理》课程教学资源（PPT课件，英文）Chapter 9 Rotational dynamics

Chapter 9 Rotational ynamics

Chapter 9 Rotational dynamics

9-1 Torque 1,Torque In this chapter we will consider only case in which the rotational axis is fixed in z M direction Fig 9-2 shows an P arbitrary rigid body that is free to rotate about the z axis A force F is applied at point P, which is located a perpendicular distance r from the axis of rotation. Fand r lie in x-y plane, and make an angle 0

9-1 Torque 1. Torque In this chapter we will consider only case in which the rotational axis is fixed in z direction. Fig 9-2 shows an arbitrary rigid body that is free to rotate about the z axis. P z O M  F  r   d M  * A force is applied at point P, which is located a perpendicular distance r from the axis of rotation. and lie in x-y plane, and make an angle . → F → F → r 

The radial component Fp= Fcos0 has no effect on rotation of the body about z axis Only the tangential component f,= Fsin Produces a rotation about the z axis The angular acceleration also depends on the magnitude of r The rotational quantity torque"T is defined as t=rF sin 0 (9-1) The unit of torque is the Newton-meter (Nm

The radial component has no effect on rotation of the body about z axis. Only the tangential component produces a rotation about the z axis. FR = F cos F⊥ = F sin  The angular acceleration also depends on the magnitude of r The rotational quantity “torque” is defined as (9-1) The unit of torque is the Newton-meter ( )   = rF sin  N m

When T =0 T=rF sin e If r=o-that is the force is applied at or through the axis of rotation If 8=or 180, that is the force is applied in the radial direction; If F=0

If r=0 -that is the force is applied at or through the axis of rotation; If or , that is the force is applied in the radial direction; If =0.   = 0 180 When =0 ?   = rF sin  → F

2.Torque as a vector t=rEsin e In terms of the cross product the torque Is expressed as 三F F (9-3) Magnitude of rEsin e Direction of t: using righ-hand rule O P

2.Torque as a vector In terms of the cross product, the torque is expressed as (9-3) → → →  = r F  = rF sin  → Magnitude of :  Direction of : using righ-hand rule → rF sin   P z O M  F  r   d M  *

Components of torque xi+yj+zk → F=Fi+f i+Fk k Sor=r×F ∧ (yF -EFi+(=F -xFj+(xFy-yFrk

      → → → = − + − + − =  = yF zF i zF x F j x F yF k F F F x y z i j k r F z y x z y x x y z ( ) ( ) ( )  →    r = x i+ y j+ z k →    F = F i+ F j+ F k x y z Components of torque: So

Sample problem 9-1 Fig 9-5 shows a Fig 9-5 pendulum The magnitude of the torgue due to gravity about the point o is I=Lmg sin 6 it has the opposite mg mg direction

Sample problem 9-1 Fig 9-5 shows a pendulum . The magnitude of the torque due to gravity about the point o is it has the opposite direction. Fig 9-5  = Lmg sin    mg mg x

9-2 Rotational inertia and newtons second law 1. Rotational inertia of a single particle Fig 9-7 shows a Fig 9-7 single particle of mass m which is attached by a thin rod of length Fsin, 6 r and of negligible mass, and free rotates about the z axis

9-2 Rotational inertia and Newton’ s second law 1. Rotational inertia of a single particle Fig 9-7 shows a single particle of mass m which is attached by a thin rod of length r and of negligible mass, and free rotates about the z axis. o x y r m Fsin  → F Fig 9-7

A force f is applied to the particle in a direction at an angle 0 with the rod(in x-y plane) Newton's Second law applied the tangential motion of the particle gives F=ma,=Fsin8 and a. =ar we obtain fsin 0=ma r rEsin e=rma.r F na s t=la I=mr

A force is applied to the particle in a direction at an angle with the rod(in x-y plane). Newton’s Second law applied the tangential motion of the particle gives and , we obtain → F  Ft = mat = F sin  F m r  =  z sin a r t =  z Fsin m r  = z r r z mr  z  2 = → → F = m a 2  = I ,I = mr z z

We define mr2 to be the " rotational inertia I of the particle about point o l= mr (9-6) The rotational inertia depends on the mass of the particle and on the perpendicular distance between the particle and the axis of rotation

We define to be the “rotational inertia” I of the particle about point o. (9-6) The rotational inertia depends on the mass of the particle and on the perpendicular distance between the particle and the axis of rotation. 2 I = mr 2 mr