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