Angular Momentum of a Ball at Half Height

And note that this is the speed of the mass as it passes through the half-way point on its way back down. It's not the final speed right at the ground.In summary, the conversation discusses the calculation of angular momentum and torque for a ball shot directly upward with initial speed 58.3 m/s. The angular momentum is found to be 0 at maximum height and -22.43934157 kg*m^2/s at half height. The torque on the ball is found to be correct at maximum height and halfway back to the ground. The conversation also addresses a sign error in the calculation of the ball's speed at half height, which should be positive rather than negative.
  • #1
ScreamingIntoTheVoid

Homework Statement


In the figure, a 0.400 kg ball is shot directly upward at initial speed 58.3 m/s. What is its angular momentum about P, 1.42 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


v^2=(v0)^2 -2gx
L=m(r x v)
J= F x r

The Attempt at a Solution


I got a,c, and d right (I'm including them just in case some other confused soul passes by and needs a hint), but I couldn't get b.

a) 0 because velocity is 0
b)
Max hight: v^2=(v0)^2 -2gx --> 0 = (58.3^2) -2(9.8)x --> 173.4127551 m -->divide by 2 to get half height = 86.70637755 m
again, v^2=(v0)^2 -2gx --> v^2=0 - 2(9.8)(86.70637755) --> -39.50588305 m/s [assuming the math is right]
L=mrv --> (0.4kg)(-39.505m/s)(1.42m) --> -22.43934157 kg*m^2/s [This is wrong even if I just do magnitude]
c and d) f x r=J --> (fx)(ry)-(fy)(rx)= J ~> (got right so the answer doesn't matter)

Can anyone tell me what went wrong in problem b? Thanks in advance to anyone who responds
 
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  • #2
ScreamingIntoTheVoid said:
v^2=0 - 2(9.8)(86.70637755)

Note that the left side is positive (it's the square of a number) while the right hand side is negative. So, something's wrong here. Once you identify the source of the sign error, check the calculation.
 
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  • #3
TSny said:
Note that the left side is positive (it's the square of a number) while the right hand side is negative. So, something's wrong here. Once you identify the source of the sign error, check the calculation.
Oh, your right if I left it there I'd get an imaginary root rather than the -39. Would V0 be what I'm looking for rather than V or is there something else I'm not noticing that's off?
 
  • #4
ScreamingIntoTheVoid said:
Oh, your right if I left it there I'd get an imaginary root rather than the -39. Would V0 be what I'm looking for rather than V or is there something else I'm not noticing that's off?
You need to think about the sign of x in your equation v^2=(v0)^2 -2gx. You set V0 = 0. So, I think that means that you are considering the mass falling back down from its maximum height. But as it falls, the displacement x it downward.

Use your calculator to check your calculation of the number 39.
 
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  • #5
TSny said:
You need to think about the sign of x in your equation v^2=(v0)^2 -2gx. You set V0 = 0. So, I think that means that you are considering the mass falling back down from its maximum height. But as it falls, the displacement x it downward.
Ok, that makes sense. So instead it would be v^2=0 +2(9.8)(86.70637755) --> 39.50588305. That'd still get me 22.43934157 in the end though.
 
  • #6
ScreamingIntoTheVoid said:
Ok, that makes sense. So instead it would be v^2=0 +2(9.8)(86.70637755) --> 39.50588305. That'd still get me 22.43934157 in the end though.
You must be making a careless error in crunching out the number 39.5
 
  • #7
You're right, got 41.2 this time, I suppose I just suck at sticking things into my calculator. Thanks!
 
  • #8
Good work. We all make such errors.
 
  • #9
1. When you calculate v at half height, since you are using v0 = 0 at the top, the x in the formula for v2 should be negative. So you will get v2 positive. I am getting the value for v slightly higher than you got so you should check that.

2. You used the components of F and r, when you calculated the torque. Why did you not do the same thing when you calculated the angular momentum? Use the correct rx, ry, vx, vy and you will get the answer.
 
  • #10
41.2 m/s is correct for the speed at half height.
 

Related to Angular Momentum of a Ball at Half Height

1. What is the Angular Momentum ball problem?

The Angular Momentum ball problem is a physics problem that involves calculating the angular momentum of a ball rolling down an incline plane. This problem is commonly used to illustrate the principles of angular momentum and rotational motion.

2. What are the key concepts involved in solving the Angular Momentum ball problem?

The key concepts involved in solving the Angular Momentum ball problem are angular momentum, rotational motion, torque, and conservation of energy. These concepts are used to determine the initial and final angular momentum of the ball, and to calculate the angular velocity and acceleration of the ball as it rolls down the incline plane.

3. How do you calculate the angular momentum of the ball in the Angular Momentum ball problem?

The angular momentum of the ball can be calculated using the formula L = Iω, where L is the angular momentum, I is the moment of inertia of the ball, and ω is the angular velocity of the ball. The moment of inertia can be calculated using the formula I = mr², where m is the mass of the ball and r is the distance from the axis of rotation to the ball.

4. How is the principle of conservation of energy applied in solving the Angular Momentum ball problem?

In the Angular Momentum ball problem, the principle of conservation of energy is applied to determine the final angular velocity and acceleration of the ball. The initial potential energy of the ball at the top of the incline plane is converted into kinetic energy as it rolls down the plane. This energy is conserved, so the final kinetic energy of the ball can be calculated and used to determine its final angular velocity and acceleration.

5. What are some real-life applications of the Angular Momentum ball problem?

The Angular Momentum ball problem has many real-life applications, such as in sports and engineering. For example, it can be used to calculate the angular momentum and trajectory of a thrown ball in sports like baseball and football. In engineering, the principles involved in solving this problem are used in designing and optimizing rotating machinery and vehicles.

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