To study torque experimentally, you apply a force to a beam. One end of the beam is attached to a pivot that allows the beam to
rotate freely. The pivot is taken as the origin or your coordinate system. You apply a force of F = Fx i + Fy j + Fz k at a point r = rx i + ry j + rz k on the beam.Part (a) Enter a vector expression for the resulting torque, in terms of the unit vectors i, j, k and the components of F and r. Part (b) Calculate the magnitude of the torque, in newton meters, when the components of the position and force vectors have the values rx = 4.07 m, ry = 0.075 m, rz = 0.035 m, Fx = 2.8 N, Fy = 8.4 N, Fz = 1.4 N. Part (c) If the moment of inertia of the beam with respect to the pivot is I = 241 kg˙m², calculate the magnitude of the angular acceleration of the beam about the pivot, in radians per second squared.
Assumption: the air resistance on this ball is negligible. Take .
a. The momentum of the ball would be approximately two seconds after it is tossed into the air.
b. The momentum of the ball would be approximately three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.
Explanation:
The momentum of an object is equal its mass times its velocity . That is: .
Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant . In other words, its velocity would become approximately more negative every second.
The initial velocity of the ball is . After two seconds, its velocity would have become . The momentum of the ball at that time would be around .
When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant . That's how the ball's velocity becomes negative.
After three more seconds, the velocity of the ball would be . Accordingly, the ball's momentum at that moment would be .