In the figure below, a 0.400 kg ball is shot directly upward at initial speed 42

Question

In the figure below, a 0.400 kg ball is shot directly upward at initial speed 42.5 m/s. What is its angular momentum about P, 2.20 m horizontally from the launch point, when the ball is at the following heights? its maximum height halfway back to the ground What is the torque on the ball about point P due to the gravitational force when the ball is at the following heights? its maximum height halfway back to the ground Angular momentum is the cross product of a position vector (extending from a chosen point, here the origin, to the particle) and a momentum vector. Do you recall that momentum is p = mv? Do you recall how to calculate velocity for the constant-velocity situation of projectile flight? Torque is the cross product of a position vector (extending from a chosen point to the particle) and a force vector.

Explanation / Answer

c. and d.) The torque is the same at the top and halfway down because the perpendicular distance from the force vector to the launch point is always 2.20 meters. The magnitude of the torque = m g (2) = 0.4*10/2=2 Nm

For a), v was obviously 0 at max height, so the angular momentum was 0 as well.

b.)First, I needed to find the halfway position in regards to y, to help find the velocity there

0 ^2 = 42.5^ 2 + 2 ( 9.8 ) ( y 0 ) , y = 92.1 m

Now using the same equation, except with vfinal being halfway up and y obviously half of what it was: v^ 2 = 0^ 2 + 2 ( 9.8 ) ( 46 0 ) ,v= 30 m/s

Now, ang mom = 30*2.20*0.4= 26.4 m/s

Reference https://www.physicsforums.com/threads/angular-momentum-of-shot-question.546418/

Reference https://www.physicsforums.com/threads/angular-momentum-of-shot-question.546418/

Reference https://www.physicsforums.com/threads/angular-momentum-of-shot-question.546418/

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