Irodov 1.94: Gravity Never Touches the Push – Amit & Deepali Sinha Foundation

Problem Irodov 1.94

A particle of mass $m$ moves on the inside of a smooth vertical cylindrical wall of radius $R$ , touching the wall throughout its motion. At the initial instant its speed is $v_0$ , directed at angle $\alpha$ above the horizontal. Find the force the particle exerts on the wall.

Side view at the initial instant. The particle sits on the inside of the vertical wall, and its velocity v_0 makes the angle α with the horizontal, resolved into v_0α across the wall and v_0α up the axis. — Side view at the initial instant. The particle sits on the inside of the vertical wall, and its velocity $\vect v_0$ makes the angle $\alpha$ with the horizontal, resolved into $v_0\cos\alpha$ across the wall and $v_0\sin\alpha$ up the axis.

Given

$m$
$R$
$v_0$
$\alpha$ (angle of $\vect v_0$ above the horizontal)

Find

the force the particle exerts on the wall

Solution

Step 1Splitting the launch velocity

Resolve the initial velocity into a component tangent to the cylinder’s horizontal cross-section and a component along the vertical axis.

v_{h}(0) = v_0\cos\alpha, \qquad v_{z}(0) = v_0\sin\alpha.

Step 2The horizontal speed never changes

The wall is smooth, so its reaction $\vect N$ on the particle is purely radial (horizontal) at every instant, and gravity contributes nothing horizontal either; the only horizontal acceleration $\vect a_h$ is therefore always perpendicular to the horizontal velocity $\vect v_h$ , which stays tangent to the fixed circle of radius $R$ .

\frac{d}{dt}\left(v_h^2\right) = 2\,\vect v_h\cdot \vect a_h = 0 \quad\text{(}\vect a_h\perp\vect v_h\text{ at every instant).}

v_h = v_0\cos\alpha \quad\text{for all } t

(1)

Step 3Newton’s second law for the horizontal motion

Top view at any later instant: the particle A runs round the circle of radius R about the axis O at the constant horizontal speed v_0α, with the wall's radial reaction N pointing inward. — Top view at any later instant: the particle $A$ runs round the circle of radius $R$ about the axis $O$ at the constant horizontal speed $v_0\cos\alpha$ , with the wall’s radial reaction $\vect N$ pointing inward.

With $v_h$ fixed at $v_0\cos\alpha$ by (1), the horizontal motion is uniform circular motion of radius $R$ , so the wall’s normal reaction supplies exactly the centripetal force for that speed.

N = \frac{mv_h^2}{R} = \frac{m\left(v_0\cos\alpha\right)^2}{R}. \tag{2}

Step 4The force on the wall

By Newton’s third law the particle pushes on the wall with a force $F$ of the same magnitude as the wall’s reaction $N$ on the particle, directed outward, so (2) carries over unchanged.

F = N.

F = \dfrac{mv_0^2}{R}\cos^2\alpha

(3)

Check

Dimensionally $[mv_0^2/R] = \text{kg}\cdot(\text{m/s})^2/\text{m} = \text{N}$ , a force. At $\alpha=0$ (launched horizontally, tangent to the wall) it reduces to the plain circular-motion result $mv_0^2/R$ ; at $\alpha=90^\circ$ (launched straight up the axis, with no horizontal component at all) it gives $F=0$ , consistent with a particle that travels straight up a single generator line of the cylinder and never curves at all.

Remark

Because the wall is smooth and vertical, its reaction is always horizontal while gravity is always vertical, so the two accelerations never project onto each other: the vertical motion (free fall, $v_z(t)=v_0\sin\alpha-gt$ ) and the horizontal motion (uniform circular motion) run completely independently, and $F$ stays constant for the whole flight even though the particle’s helical path keeps changing its vertical speed. A curved bowl or a sloped wall would not have this property, since its reaction would then have a vertical component that varies as the particle’s height on the surface changes.