Fall 2003 1 Mech

A homogeneous ball of mass $M$ and radius $R$ is struck impulsively at its center, causing it to go from rest to a horizontal speed of $v_0$. Assuming a constant coefficient of friction $\mu$, find the distance traveled by the ball before it begins to roll without slipping.

**

Answer

**

The requirement for rolling without slipping is $\omega R = v_{x,cm}$. We need to write down the equations of motion for $x_{cm}$, $y_{cm}$, the position of the center of mass, and $\theta$, the total angle the ball has rolled, and solve. Applying all forces and torques, we get for the $y$, $x$, and $\theta$ motions

(1)
\begin{eqnarray} -M g + N &=& M a_y = 0 \\ -\mu N &=& M a_x \\ R \mu N &=& I \alpha = \frac{2}{5}M R^2 \alpha \end{eqnarray}

Solving for $v_x$, we get

(2)
\begin{align} v_x = v_0 - \mu g t \end{align}

Solving for $\omega$, we get

(3)
\begin{align} \omega = \frac{5 \mu g}{2 R} t \end{align}

Setting $\omega R = v_x$ at time $T$, we get

(4)
\begin{eqnarray} \frac{5 \mu g}{2} T &=& v_0 - \mu g T \\ \Rightarrow T &=& \frac{2 v_0}{7 \mu g} \end{eqnarray}

To find the distance it traveled in this period, we must find the position of the center of mass at this time

(5)
\begin{align} x_{cm} = -\frac{1}{2} \mu g t^2 + v_0 t \end{align}

at time $T$, this is

(6)
\begin{align} x(T) = \frac{12}{49} \frac{v_0^2}{\mu g} \end{align}
Add a New Comment
or Sign in as Wikidot user
(will not be published)
- +
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-Share Alike 2.5 License.