Question

I got A but b,c,d are wrong

In the figure below, a 0.390 kg ball is shot directly upward at initial speed 41.5 m/s.

What is its angular momentum about P, 1.85 m horizontally from the launch point, when the ball is at thefollowing heights?(a) its maximum height
kg·m²/s

(b) halfway back to the ground
kg·m²/sWhat is the torque on the ball about point P due to the gravitational force when the ball is at the following heights?(c) its maximum height
N·m

(d) halfway back to the ground
N·m

Answer 1

a)Angular momentum is proportional to velocity. at the maximum height, it will have 0 velocity so:

L=0 kg-m²/s

b)At half the max height the kinetic energy will only be half so the velocity will only be√(1/2)*41.5

L=mvr=.39(√(1/2)*41.5)(1.85)

L=21.17 kg-m²/s

c)Torque is the product of the force and perpendicular radius. Because the ball is going straight up the perpendicular radius doesn't change. Force ofgravity is also constant, so the torque doesn't change with vertical position.

T=FR=mgr=.39(9.8)(1.85)

T=7.0707 N-m

d) The same as above because constant with all vertical positions

T=7.0707 N-m

Hope that helps