Question
please help
The position vector of a particle of mass m at time t is r = R [cos (omega t) I + sin (omega t) j]. where R and omega are positive constants. In terms of r the net force exerted on the particle is given by A particle of mass m is moving with constants velocity vi along the straight line y - b in the xOy plane, where b is a positive constant. The angular momentum of the particle at point (x, b) with respect to the coordinate origin is A uniform cube of side length a and mass M is lifted from a horizontal floor so that it rotates about an edge that remains in full contact with the floor and does not slide. In the lifting process, the maximum magnitude of the torque exerted on the cube about the edge by the weight of the cube is To appear to be thin observers at rest, a young man starts a spacecraft. To reduce his thickness exactly by half, the speed at which the spacecraft must move is
Explanation / Answer
1) v = dr/dt = RW(-sinwti+coswtj)
a = dv/dt =- RW^2(COSWTI+SINWTJ)
f = ma = -mRw^2(coswti+sinwtj)
2) L = mvi*bk = mbv(i*k) = -mvbj
3) torque = r*f = *a/sqrt2)*mg = mga/1.414
4) l = l0*sqrt(1-v^2/c^2)
l0/2 = lo*sqrt*1-v^2/c^)
1-v^2/c^2 = 1/4
v^2/c^2 = 3/4
v/c = 0.866
v = 0.866c
