How Do You Correctly Cancel Angular Momentum in Physics Problems?

  • #1
pbnj
5
0
Homework Statement
A ball with mass ##m## and diameter ##D## is thrown with speed ##v## at an angle ##\theta## with the horizontal from a height ##h_i##. How much spin (in rad/s) must the thrower impart on the ball so that at its maximum height, it has no angular momentum with respect to a point on the ground directly beneath the ball?
Relevant Equations
##v_f^2 = v_i^2 + 2a\Delta h##
##L = I\omega##
##L = \vec r \times \vec p##
First, we calculate the maximum height using the first equation, noting that at maximum height, the velocity is purely horizontal with speed ##v\cos\theta##, and with initial vertical speed ##v\sin\theta##:

$$
\begin{align}
v_f^2 &= v_i^2 - 2g(h_f - h_i) \\
0 &= (v\sin\theta)^2 - 2g(h_f - h_i) \\
h_f &= \frac{(v\sin\theta)^2}{2g} + h_i
\end{align}
$$

The angular momentum from translation is given by ##L_t = \vec r \times \vec p = h_f\hat j \times mv\cos\theta\hat i = -h_fmv\cos\theta\hat k##. The moment of inertia of a sphere is ##I = \frac 2 5 mR^2## where ##R = \frac D 2##, and its angular momentum due to spin is ##L_s = I\omega##.

We need these to cancel, so

$$
\begin{align}
L_t + L_s &= 0 \\
-h_fmv\cos\theta + \frac 2 5 mR^2\omega &= 0 \\
\omega &= \frac{h_fmv\cos\theta}{\frac 2 5 mR^2} \\
&= \frac{5h_fv\cos\theta}{2R^2}
\end{align}
$$

The problem uses ##m = 625g##, ##D = 22.9cm##, ##\theta = 45^\circ##, ##v=5m/s## and ##h_i = 1.5m##. Using these values I get a spin of about ##1440.82##, in rad/s.
 
Physics news on Phys.org
  • #2
Your answer agrees with mine.
 
  • #3
pbnj said:
Using these values I get a spin of about 1440.82, in rad/s.
Have you a reason to doubt it?
 
  • #4
haruspex said:
Have you a reason to doubt it?
It's for a Coursera course, and it tells me it's incorrect. I've tried incrementing/decrementing my answer by 5 a few times, I tried using 2 sigfigs for everything, I tried assuming "diameter" meant "radius," but no luck. On the discussion forum there are no questions about this particular answer (the course is 3 years old, and it doesn't seem like new posts get replies). If my approach is correct and my calculations are correct, I can only assume something in the backend changed in the 3 years between then and now.
 

Similar threads

  • Introductory Physics Homework Help
2
Replies
55
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
665
  • Introductory Physics Homework Help
Replies
10
Views
312
  • Introductory Physics Homework Help
Replies
7
Views
246
  • Introductory Physics Homework Help
Replies
6
Views
598
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
236
  • Introductory Physics Homework Help
Replies
15
Views
331
  • Introductory Physics Homework Help
10
Replies
335
Views
8K
  • Introductory Physics Homework Help
Replies
13
Views
2K
Back
Top