A force `vecF=(3hati+4hatj)N` is acting on a point mass `m=((1)/(2))kg` at a point `A(2m,2m)`. Find the angular acceleration of the line `OA` at this instant.

A force F=(2hati+3hatj-2hatk)N is acting on a body at point (2m, 4m,-2m) . Find torque of this force about origin.

A force F=(2hati+3hatj-2hatk)N is acting on a body at point (2m, 4m,-2m) . Find torque of this force about origin.

A uniform rod of mass M and length L is free to rotate in X-Z plane i.e. vecF=(3hati+2hatj+6hatk)N is acting on the rod at (L//2, 0,0) in the situtation shown in figure. The angular acceleration of the rod is (Take M=6 kg and L=4m )

A uniform rod of mass M and length L is free to rotate in X-Z plane i.e. vecF=(3hati+2hatj+6hatk)N is acting on the rod at (L//2, 0,0) in the situtation shown in figure. The angular acceleration of the rod is (Take M=6 kg and L=4m )

A force vecF=4hati-5hatj+3hatk N is acting on a point vecr_1=2hati+4hatj+3hatk m. The torque acting about a point vecr_2=4hati-3hatk m is

A force vecF=4hati-5hatj+3hatk N is acting on a point vecr_1=2hati+4hatj+3hatk m. The torque acting about a point vecr_2=4hati-3hatk m is

A force vecF=(hati+3hatj)N acts on a body at a point P, which is at a distance given by vecr=(3hati+hatj)m from the axis of rotation. What is the direction of the torque acting on the body?