Solved Problem on Two-dimensional Motion and Relative Motion

Solved Problem on Relative Motion

A worker holds one end of a straight board of length a, while the other end rests on a cylindrical drum so that the board is in a horizontal position. When moving the board forward, the worker makes the drum roll without slipping along the horizontal plane, and during the movement, the board remains horizontal. Determine the distance d that the worker will travel until the end he is holding touches the drum.

Problem data:

Board length: a.

Solution

Let's call \( {\vec v}_{\small C} \) the speed of the center of the drum relative to the ground, \( {\vec v}_{\small{B/C}} \) the speed of the point of contact between the drum and the board relative to the center of the drum, and \( {\vec v}_{\small B} \) the speed of the point of contact between the drum and the board relative to the ground (Figure 1). Taking v for the velocity of the center of the drum, the velocity of the point of contact relative to the center will also be v (if this velocity were higher or lower the drum would deform), then we can calculate the velocity of the point of contact relative to the ground

Figure 1

\[ \begin{gather} {\vec v}_{\small B}={\vec v}_{\small C}+{\vec v}_{\small{B/C}} \end{gather} \]

since all vectors have the same direction, in magnitude

\[ \begin{gather} v_{\small B}=v+v\\[5pt] v_{\small B}=2v \end{gather} \]

Figure 2

As the board passes through the drum without slipping, all points on the board have the same speed of 2v as the point of contact. The point of contact of the worker with the board also has a speed of 2v, and the worker himself moves with a speed of 2v (Figure 2).

Figure 3

The problem starts with one end of the board held by the worker and the other end resting on the drum, this support point is exactly over the center of the drum (Figure 3). The problem ends when the worker's hand is in contact with the drum, at this moment it will be on the center of the drum.

Then we can "forget" the worker, the drum, and the board (Figure 4) and reduce the problem to two material point masses. We choose a frame of reference oriented to the left (contrary to what is usually done), a point represents the worker's hand starting from the origin S_0H = 0 with initial speed v_0H = v_H = 2v, and another point represents the center of the drum which starts from a point S_0C = a with speed v_0C = v_C = v. As the speeds of the points are constant we use the equation of displacement as a function of time is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {S=S_{0}+vt} \end{gather} \]

Figure 4

writing this equation for the two-point masses

\[ \begin{gather} S_{\small H}=S_{0\small H}+v_{\small H}t\\[5pt] S_{\small H}=0+2vt\\[5pt] S_{\small H}=2vt \tag{I} \end{gather} \]

\[ \begin{gather} S_{\small C}=S_{0\small C}+v_{\small C}t\\[5pt] S_{\small C}=a+vt \tag{II} \end{gather} \]

for them to meet must occupy the same position, then we must impose the condition

\[ \begin{gather} S_{\small H}=S_{\small C} \end{gather} \]

equating expressions (I) and (II) we obtain the time it takes for the points to meet

\[ \begin{gather} 2vt=a+vt\\[5pt] 2vt-vt=a\\[5pt] a=vt\\[5pt] t=\frac{a}{v} \tag{III} \end{gather} \]

substituting expression (III) into equation (I) we have the distance traveled by man

\[ \begin{gather} S_{\small H}=2\cancel v\frac{a}{\cancel v} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {S_{\small H}=2a} \end{gather} \]

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