- #1

pbnj

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- 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.

$$

\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.