# A thin rod of length 1 m is fixed in a vertical position inside a train, which is moving...

## Question:

A thin rod of length 1 m is fixed in a vertical position inside a train, which is moving horizontally with constant acceleration 4 m/s{eq}^2 {/eq}. A bead can slide on the rod, and the friction coefficient between them is 1/2. If the bead is released from rest at the top of the rod, find the time when it will reach the bottom.

## Pseudo Force

A pseudo force is a type of force that is introduced in a system that is observed in an accelerated frame of reference. Taking this force into account is necessary when solving problems that involves different reference frames for consistency.

The given for this problem are the following:

• The rod length {eq}\begin{align*} y=1~m \end{align*} {/eq}
• The uniform acceleration of the train {eq}\begin{align*} a_{t}=4\frac{m}{s^2} \end{align*} {/eq}
• The frictional coefficient {eq}\begin{align*} \mu=\frac{1}{2}=0.5 \end{align*} {/eq}
• Accleration due to gravity {eq}\begin{align*} g=9.8\frac{m}{s^2} \end{align*} {/eq}
• Solve for the time that the bead reaches the bottom.

(a) For this problem, we first solve for the vertical acceleration of the bead. This is done by calculating the net force acting on the bead in the vertical direction. The forces acting on the sliding bead is its weight {eq}\begin{align*} W=mg \end{align*} {/eq} countered by the friction {eq}\begin{align*} F_{fric}=\mu N \end{align*} {/eq} where the normal force {eq}\begin{align*} N \end{align*} {/eq} is a pseudo force. The vertical acceleration is determined to be

\begin{align*} \sum F_y&=W-F_{fric}&\text{[Expanding the terms]}\\\\ ma_{ver}&=mg-\mu N&\text{[Note: N is a pseudo force due to the acceleration of the train]}\\\\ ma_{ver}&=mg-\mu (ma_{t})&\text{[Cancel out the mass]}\\\\ a_{ver}&=g-\mu a_{t}&\text{[Substitute the required values]}\\\\ a_{ver}&=9.8-0.5\left( 4\right)&\text{[Simplify]}\\\\ a_{ver}&=\boxed{7.8\frac{m}{s^2}} \end{align*}

(b) With the obtained acceleration in part (a), we can use the equation of motion in the vertical direction. This is determined to be

\begin{align*} y&=y_0t+\frac{1}{2}a_{ver}t^2&\text{[Recall that the initial position is zero]}\\\\ 1&=0+\frac{1}{2}(7.8)t^2&\text{[Rearrange]}\\\\ t&=\sqrt{\frac{(2)(1)}{7.8}}&\text{[Simplify]}\\\\ t&=\boxed{0.5063696835~s} \end{align*} 