Irodov 1.15: The Bolt That Keeps Rising First – Amit & Deepali Sinha Foundation

Problem Irodov 1.15

An elevator car has a floor-to-ceiling height $h = 2.7$ m. Starting from rest, it accelerates upward at a constant rate $a = 1.2$ m/s $^2$ . A time $t_0 = 2.0$ s after the start, a bolt comes loose from the ceiling and falls freely (under gravity alone) until it strikes the floor, while the car continues to accelerate upward at $a$ throughout. Find (a) the elapsed time between the bolt coming loose and striking the floor, and (b) the bolt’s net displacement and the total distance it travels during that fall, both measured in the frame fixed to the elevator shaft (the ground).

Position versus time, measured from the ground, for the car's ceiling and its floor. Both climb together under the constant acceleration a, staying the fixed height h apart, and the dot marks the instant t_0 at which the bolt comes loose from the ceiling. — Position versus time, measured from the ground, for the car’s ceiling and its floor. Both climb together under the constant acceleration $a$ , staying the fixed height $h$ apart, and the dot marks the instant $t_0$ at which the bolt comes loose from the ceiling.

Given

$h = 2.7$ m
$a = 1.2$ m/s $^2$
$t_0 = 2.0$ s

Find

(a) the fall time

t_1

, (b) the displacement

\Delta x

and the distance

s

covered by the bolt, ground frame

Solution

Step 1Velocity at the instant of release

The car accelerates uniformly from rest, so integrating once gives its speed at the moment the bolt lets go; this is also the bolt’s own speed at that instant, since up to then it has been moving with the car.

v_0 = \int_0^{t_0} a\,d\tau = a t_0

Step 2Position of the bolt and of the floor after release

Let $t'$ be the time elapsed since release, and measure position from the bolt’s location at that instant. The bolt now accelerates at $-g$ (gravity only); the floor, still part of the car, keeps accelerating at $+a$ and starts a distance $h$ below the bolt. Integrating each equation of motion twice from the common initial velocity $v_0$ ,

\begin{align*} x_{\text{bolt}}(t') &= v_0 t'-\dfrac{g t'^2}{2} \tag{1} \\ x_{\text{floor}}(t') &= -h+v_0 t'+\dfrac{a t'^2}{2} \tag{2}\end{align*}

The bolt and the floor plotted from the ground. Both track the car up to release; past it the bolt bends over under gravity while the floor keeps climbing under a, and the two meet at the landing. — The bolt and the floor plotted from the ground. Both track the car up to release; past it the bolt bends over under gravity while the floor keeps climbing under $a$ , and the two meet at the landing.

Step 3Fall time from where the two positions meet

The bolt strikes the floor when $x_{\text{bolt}}(t_1)=x_{\text{floor}}(t_1)$ . Subtracting (2) from (1) and canceling the equal $v_0 t_1$ terms leaves

\begin{align*}v_0 t_1-\dfrac{g t_1^2}{2} &= -h+v_0 t_1+\dfrac{a t_1^2}{2} \\ h &= \dfrac{(a+g) t_1^2}{2}\end{align*}

t_1 = \sqrt{\dfrac{2h}{a+g}} = \sqrt{\dfrac{2(2.7)}{1.2+9.8}} = 0.70\text{ s}

(3)

Step 4Net displacement of the bolt

Substituting (3) for $t_1$ into the bolt equation (1), using $t_1^2=2h/(a+g)$ to replace the quadratic term, gives the bolt’s ground-frame displacement over the fall.

\Delta x = v_0 t_1-\dfrac{g t_1^2}{2}

Numerically,

\Delta x = (1.2)(2.0)(0.7006)-\dfrac{(9.8)(2.7)}{11.0}\text{ m}

\Delta x = v_0 t_1-\dfrac{gh}{a+g} = -0.72\text{ m}

(4)

negative, so the bolt ends up $0.72$ m below its release point.

Step 5Total distance covered, up then down

The bolt’s velocity $v_0-gt'$ vanishes at the apex time

t_a = \dfrac{v_0}{g} = \dfrac{2.4}{9.8} \approx 0.24\text{ s},

which is less than the fall time $t_1$ of (3), so the bolt rises to an apex before descending; the path length below is therefore the climb to the apex plus the drop from the apex to the landing point, not the straight-line displacement.

\begin{align*}x_{\text{apex}} &= \dfrac{v_0^2}{2g} \\ s &= x_{\text{apex}}+\left(x_{\text{apex}}-\Delta x\right)\end{align*}

s = \dfrac{v_0^2}{g}-\Delta x = \dfrac{(2.4)^2}{9.8}-(-0.72) = 1.3\text{ m}

(5)

Check

Dimensionally, $\sqrt{h/(a+g)}$ is a time and $v_0^2/g$ a length, as required. If the car were not accelerating ( $a=0$ ) then $v_0=0$ too, and (3) collapses to the textbook drop time $t_1=\sqrt{2h/g}$ with $\Delta x=-h$ and $s=h$ , the ordinary free fall of a bolt released from rest, exactly as it should.

Remark

Inside the car the bolt simply falls the height $h$ in $0.70$ s, but relative to the ground it does something less obvious: since it keeps its upward release velocity $v_0$ , it first rises further before gravity turns it around, so the path length $s$ exceeds the net drop $|\Delta x|$ by the round trip to the apex and back, $v_0^2/g$ , not by a coincidence of the numbers. An accelerating frame can hide real motion, not just add a fictitious force to explain it away.