We established Newton\'s Second Law for rotations as t rightarrow_net = I alpha

Question

We established Newton's Second Law for rotations as t rightarrow_net = I alpha rightarrow We want to determine if Newton's Third Law can be applied to rotations about a single axis. Imagine you have two point objects, a and b located at points (x_1, y_1) and (x_2, y_2), respectively, in a given Cartesian coordinate system. The two objects exert forces F rightarrow_a rightarrow b and F rightarow_b rightarrow a = - Frightarrow_a rightarrow b on each other and the forces are directed along the line joining the two objects. It should be clear if x_1 = x_2 and y_1 = y_2 the two oppositely directed forces are exerted at the same location. Since r rightarrow is the same and the magnitudes of the forces are the same, they will produce equal and opposite torques about any chosen axis and Newton's Third Law for rotations applies for contact forces. But what about non-contact forces, like gravity? Assume x_1 not equal to x_2 and y_1 not equal to y_2. From T = T r rightarrow times F rightarrow and using the origin for the axis of rotation determine the torque on a due to the force from b. Next, for the same axis of rotation, determine the torque on b due to the force from a. Finding the torques may be accomplished using either T = rF sin theta or via the determinant form. If you use the angle you will need to rely on your Euclidean geometry skills. If you use the determinant form you will need to rely on your coordinate geometry skills. Based on your results, is there a rotational analogue for Newton's Third Law? Comment on this result and its implications for solving problems.

Explanation / Answer

let magnitude of each force be F.

torque of force applied by b at a about origin:

position vector r=(x1,y1)

vector along F(b-->a) =(x1,y1)-(x2,y2)=(x1-x2,y1-y2)

magnitude of vector=d=sqrt((x1-x2)^2+(y1-y2)^2)

force vector =Fba=F*(x1-x2,y1-y2)/d

then torque=cross product of r and Fba

= determinant of the matrix

i       j       k
x1       y1       0
F*(x1-x2)/d   F*(y1-y2)/d   0

=(F/d)*(x1*y1-x1*y2-x1*y1+x2*y1) k

=(F/d)*(x2*y1-x1*y2) k

torque of force applied by a at b about origin:

position vector r=(x2,y2)

vector along F(a-->b) =(x2,y2)-(x1,y1)=(x2-x1,y2-y1)

magnitude of vector=d=sqrt((x1-x2)^2+(y1-y2)^2)

force vector =Fab=F*(x2-x1,y2-y1)/d

then torque=cross product of r and Fab

= determinant of the matrix

i       j       k
x2       y2       0
F*(x2-x1)/d   F*(y2-y1)/d   0

=(F/d)*(x2*y2-x2*y1-x2*y2+x1*y2) k

=(F/d)*(-x2*y1+x1*y2) k

so total torque=(F/d)*(x2*y1-x1*y2) k + (F/d)*(-x2*y1+x1*y2) k

=0

hence the forces exert equal and opposite torques even when it is not contact force.

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