Angular momentum of ball problem

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Homework Help Overview

The problem involves calculating the angular momentum of a ball shot vertically upward and its torque about a point located horizontally from the launch point. The discussion focuses on the ball's behavior at maximum height and halfway back to the ground, as well as the effects of gravitational force.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the angular momentum formula but expresses uncertainty about their calculations, particularly at the halfway point. Participants question the accuracy of the height calculations and the signs used for acceleration due to gravity.

Discussion Status

Participants are actively engaging in clarifying the original poster's calculations, pointing out potential errors in the approach. Some guidance has been offered regarding the relationship between velocity and distance, and the original poster indicates they will attempt the problem again based on the feedback received.

Contextual Notes

There are indications of confusion regarding the signs used in the calculations, particularly relating to gravitational acceleration and initial velocity. The original poster's calculations led to results that were inconsistent with physical expectations, prompting further examination of assumptions.

J-dizzal
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Homework Statement


In the figure, a 0.400 kg ball is shot directly upward at initial speed 40.4 m/s. What is its angular momentum about P, 6.65 m horizontally from the launch point, when the ball is (a) at maximum height and (b) halfway back to the ground? What is the torque on the ball about P due to the gravitational force when the ball is (c) at maximum height and (d) halfway back to the ground?

Homework Equations


p=mv
l=r x p

The Attempt at a Solution


20150713_190337_zpsmmn73rsx.jpg

im stuck at part b, i used l=rxp but not sure if i did that right.
 
Last edited by a moderator:
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Overall, your method looks good. However, your calculation of hmax is incorrect due to a sign error in the acceleration. A clue that something's wrong is that you got a longer time to fall halfway back down than the total time to reach maximum height. Also, your speed at half the max height is greater than the launch speed.

It might be easier to relate velocity directly to distance using vf2 = vi2 +2ay.
 
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You've gone wrong right near the start in calculating max height, because you've given g the same sign as the initial velocity, whereas in fact they are opposite. So your max height will be much lower.

There's a pretty big clue to this in that you calculate the speed when the ball's halfway back down as 49+ m/s, which is more than the initial speed, which is clearly impossible.

See if fixing up the errors helps you answer the question.
Edit: Jinx TSny
 
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thank you both. ill try again.

edit. that worked thanks!
 
Last edited:
@ andrewkirk. What is that saying about certain minds thinking alike?
 

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