Irodov 1.281: The Speed That Frees One Bearing – Amit & Deepali Sinha Foundation

Problem Irodov 1.281

A uniform rod $AB$ of mass $m = 1.40$ kg and length $l_0 = 100$ cm is held horizontal and spins about a fixed vertical shaft $OO'$ that passes through the rod’s end $A$ . The shaft is carried in bearings at its two ends, the lower one $O$ and the upper one $O'$ , and its total length is $l = 55$ cm, with the attachment point $A$ exactly halfway between the bearings. Friction is absent. Find the angular velocity of the rod at which the bearing at the lower end $O$ exerts no horizontal force at all, and find the horizontal force exerted at the upper end $O'$ at that angular velocity.

The arrangement. The vertical shaft OO' turns in two bearings a distance l apart, the rod AB of length l_0 is fixed to it at A, which sits midway between the bearings, and the whole assembly spins about the shaft with angular velocity ω. — The arrangement. The vertical shaft $OO'$ turns in two bearings a distance $l$ apart, the rod $AB$ of length $l_0$ is fixed to it at $A$ , which sits midway between the bearings, and the whole assembly spins about the shaft with angular velocity $\vect\omega$ .

Given

$m = 1.40$ kg
$l_0 = 100$ cm
$l = 55$ cm

Find

the angular velocity

\omega

at which the horizontal force at

O

vanishes, and the horizontal force

F'

O'

then

Solution

Step 1The assembly’s angular momentum never changes

Take the fixed point $A$ as origin, with $z$ up along the shaft and $x$ along the rod at the instant considered. All of the rod’s mass lies on the $x$ axis, so the products of inertia at $A$ vanish and $x$ , $y$ , $z$ are principal axes there; the angular momentum about $A$ is therefore parallel to $\vect\omega$ and fixed in the laboratory.

\vect L_A = I_{zz}\,\vect\omega = \frac{m l_0^2}{3}\,\omega\,\vect k

Its magnitude and its direction are both constant in time, so its derivative vanishes, and with it the total external moment about $A$ .

\vect N_A = \frac{d\vect L_A}{dt} = \vect 0

(1)

Step 2The moment balance about

A

fixes the upper reaction

The figure below is the free body of the rotating assembly: the weight $m\vect g$ at the rod’s center, the horizontal bearing force $\vect F'$ at $O'$ , and no horizontal force at $O$ , which is the condition the problem imposes. The vertical forces at $O$ and $O'$ act along the line $AO$ , so they have no moment about $A$ and never enter.

Free body of the assembly in the state asked for: the weight at the rod's midpoint, a distance l_0/2 from the shaft; the upper bearing's force F' at height l/2 above A; the centripetal acceleration w_C. No horizontal force at the lower bearing. — Free body of the assembly in the state asked for: the weight at the rod’s midpoint, a distance $l_0/2$ from the shaft; the upper bearing’s force $\vect F'$ at height $l/2$ above $A$ ; the centripetal acceleration $\vect w_C$ . No horizontal force at the lower bearing.

Writing (1) out about $A$ , the weight’s moment arm is the horizontal distance $l_0/2$ to the center of mass and the reaction’s moment arm is the height $l/2$ of the upper bearing, and the two moments turn the assembly in opposite senses.

F'\,\frac{l}{2} = mg\,\frac{l_0}{2}

(2)

Step 3Newton’s second law supplies the same force again

The center of mass travels a circle of radius $l_0/2$ at a constant rate, so it accelerates towards the shaft with $w_C = \omega^2 l_0/2$ . With no horizontal force at $O$ , the only horizontal force on the assembly is $\vect F'$ , so it alone produces that acceleration.

F' = m w_C = m\,\omega^2\,\frac{l_0}{2}

(3)

Step 4Eliminating the reaction leaves the angular velocity

The same force appears in both (2) and (3), so substituting the second into the first removes it and leaves a condition on $\omega$ .

\begin{align*}\left(m\,\omega^2\,\frac{l_0}{2}\right)\frac{l}{2} &= mg\,\frac{l_0}{2} \\ \omega^2\,\frac{l}{2} &= g \\ \omega^2 &= \frac{2g}{l}\end{align*}

Both the mass and the rod’s length have canceled, leaving only the bearing span.

\omega = \sqrt{\frac{2g}{l}} = \sqrt{\frac{2(9.8\ \text{m/s}^2)}{0.55\ \text{m}}} = 6.0\ \text{rad/s}

(4)

Step 5The horizontal force at the upper bearing

The moment balance (2) gives that force on its own, since it never involved $\omega$ . Its direction follows from the two roles it plays at once: it closes the couple that holds the rod up against gravity, and it is the entire centripetal pull on the center of mass, so it points horizontally from the rod’s far end towards the shaft and turns with the rod. Numerically,

F' = \frac{(1.40\ \text{kg})(9.8\ \text{m/s}^2)(1.00\ \text{m})}{0.55\ \text{m}}

F' = \frac{m g l_0}{l} = 25\ \text{N}

(5)

Horizontal, pointing from

B

towards

A

, and turning with the rod.

Check

Dimensionally $\sqrt{2g/l}$ is $\sqrt{(\text{m/s}^2)/\text{m}} = 1/\text{s}$ , as an angular velocity must be. Numerically the two answers are consistent with each other: at $\omega = 5.97$ rad/s the centripetal force the center of mass demands is $m\omega^2 l_0/2 = (1.40)(35.6)(1.00)/2 = 24.9$ N, which is the same 25 N the moment balance asks of the upper bearing. Note also $F'/mg = l_0/l = 1.8$ , so the bearing carries nearly twice the rod’s weight sideways.

Remark

Solve the two conditions without imposing anything and each bearing’s horizontal force splits into two pieces: half of the couple $m g l_0/l$ that holds the rod up, which acts in opposite senses at the two ends, plus half of the centripetal pull $m\omega^2 l_0/2$ , which acts in the same sense at both. At the lower bearing the two pieces oppose each other, and the answer above is simply the spin rate at which they are equal. Spin faster and that force does not stay zero: it reappears reversed, with both bearings now pulling the shaft towards the axis, and the whole balance is set by the bearing span alone, never by how heavy or how long the rod is.