Irodov 1.115: Friction-Free Slide, Rotating Only Underneath

Problem Irodov 1.115

A small body of mass $m$ is released from rest at the very top of a smooth sphere of radius $R$ and slides down under gravity. The sphere itself spins at a constant angular velocity $\omega$ about a fixed vertical axis through its own center. Working in the reference frame fixed to the sphere, find the centrifugal force and the Coriolis force acting on the body at the instant it leaves the sphere’s surface.

The body partway down the smooth sphere, at a colatitude θ measured at the center from the crown it started at. The sphere turns at the constant angular velocity ω about the dashed vertical axis through its center. — The body partway down the smooth sphere, at a colatitude $\theta$ measured at the center from the crown it started at. The sphere turns at the constant angular velocity $\omega$ about the dashed vertical axis through its center.

Given

$m = 0.30$ kg
$R = 1.00$ m
$\omega = 6.0$ rad/s
body released from rest at the crown

Find

The centrifugal force and the Coriolis force at break-off, in the sphere’s own frame

Solution

Step 1Reference: the equation of motion in a rotating frame

In a reference frame rotating with constant angular velocity $\vect\omega$ relative to an inertial frame, a particle of mass $m$ obeys

m\vect w' = \vect F + m\omega^{2}\vect\rho + 2m\,(\vect v'\times\vect\omega)

(1)

where $\vect w'$ and $\vect v'$ are the particle’s acceleration and velocity measured in the rotating frame, $\vect F$ is the resultant of the real (interaction) forces, and $\vect\rho$ is the position vector drawn perpendicularly from the rotation axis to the particle. The term $m\omega^{2}\vect\rho$ is the centrifugal force: it depends only on where the particle is and points away from the axis. The term $2m(\vect v'\times\vect\omega)$ is the Coriolis force: it depends only on how the particle moves, is perpendicular to both $\vect v'$ and $\vect\omega$ , and has magnitude $2mv'\omega\sin\theta$ , where $\theta$ is the angle between $\vect v'$ and $\vect\omega$ . Both vanish in an inertial frame ( $\omega=0$ ), and the Coriolis force does no work, since it is always perpendicular to $\vect v'$ .

In this problem the rotating frame is the spinning sphere, so $\vect\omega$ is vertical, along the sphere’s own axis; $\vect\rho$ is horizontal, running from that axis out to the body, with magnitude $x=R\sin\theta$ rather than $R$ ; and $\vect v'$ is the body’s velocity relative to the sphere’s surface, which is not the same as its velocity $\vect v$ relative to the ground. The real forces $\vect F$ are the weight and the normal reaction, and neither of them knows about $\omega$ on a smooth sphere, so the break-off point is settled first in the inertial frame and both fictitious terms are then evaluated at that point. Only the part of $\vect v'$ perpendicular to the vertical $\vect\omega$ survives the cross product in the Coriolis term, which is why the horizontal projection of $\vect v'$ is the quantity computed below. (The letter $\theta$ names the colatitude everywhere in this solution, not the angle appearing in the reference above.)

Step 2The break-off point does not depend on the spin

The two real forces on the body at a general colatitude θ: the normal reaction N along the outward radius, and the weight m g. — The two real forces on the body at a general colatitude $\theta$ : the normal reaction $\vect N$ along the outward radius, and the weight $m\vect g$ .

The sphere is smooth, so the only real forces $\vect F$ in (1), in any frame, are gravity and the normal reaction, both drawn above; friction is the only real force that could couple the sphere’s spin to the body, and there is none. The body starts exactly on the rotation axis, where the sphere’s own surface speed is zero, so it starts from rest in the inertial frame too.

Energy conservation and the radial (centripetal) equation, in the inertial frame, with $v$ the body’s inertial-frame speed at colatitude $\theta$ ,

\frac12mv^2 = mgR(1-\cos\theta), \qquad mg\cos\theta - N = \frac{mv^2}{R}.

Step 3The break-off angle and speed

Break-off is $N=0$ . Substituting $v^2$ from the energy relation into the radial equation,

mg\cos\theta = \frac{m}{R}\bigl[2gR(1-\cos\theta)\bigr] \;\Longrightarrow\; \cos\theta = 2(1-\cos\theta).

\cos\theta = 2/3

(2)

so $\sin\theta=\sqrt{5}/3$ , and the inertial-frame speed there follows from the energy relation.

v^2 = 2gR(1-\cos\theta) = \dfrac{2gR}{3}

(3)

Step 4The centrifugal force

The body at the break-off point θ=2/3, in the sphere's rotating frame. The centrifugal force F_cf (gold) is horizontal, away from the vertical spin axis, its magnitude set by the horizontal distance x from that axis. — The body at the break-off point $\cos\theta=2/3$ , in the sphere’s rotating frame. The centrifugal force $\vect F_{cf}$ (gold) is horizontal, away from the vertical spin axis, its magnitude set by the horizontal distance $x$ from that axis.

The centrifugal term of (1) depends only on the body’s distance $x=R\sin\theta$ from the vertical spin axis, that being the magnitude of $\vect\rho$ , regardless of which frame found $\theta$ .

F_{cf} = m\omega^2x = m\omega^2R\sin\theta

Substituting $\sin\theta=\sqrt5/3$ from (2) and putting in the data,

F_{cf} = 0.30\times6.0^2\times1.00\sqrt{5/9}\text{ N}

F_{cf} = m\omega^2R\sqrt{5/9} \approx 8\text{ N}

(4)

Step 5The Coriolis force, from the horizontal part of

\vect v'

The relative velocity is $\vect v' = \vect v - \vect v_{\text{disc}}$ , where $v_{\text{disc}}=\omega R\sin\theta$ is the sphere’s own surface speed at that point, horizontal and tangent to the latitude circle. Since $\vect v$ (inertial, tangent to the meridian) and $\vect v_{\text{disc}}$ (horizontal, perpendicular to the meridian plane) are themselves perpendicular, the horizontal part of $\vect v'$ combines them in quadrature: $\vect v$ ’s own horizontal projection is $v\cos\theta$ , so

v_{\perp}'^2 = (v\cos\theta)^2+(\omega R\sin\theta)^2.

Substituting $v^2=2gR/3$ , $\cos\theta=2/3$ and $\sin\theta=\sqrt5/3$ from (2) and (3), and writing $F_{Cor}=2m\omega v_\perp'$ ,

\begin{align*}F_{Cor}^2 &= 4m^2\omega^2\left[\frac{2gR}{3}\cdot\frac49+\omega^2R^2\cdot\frac59\right] \\ &= \frac{4}{9}m^2\omega^4R^2\left[\frac{8g}{3\omega^2R}+5\right]\end{align*}

Numerically,

F_{Cor} = \dfrac23\times0.30\times6.0^2\times1.00\sqrt{5+\dfrac{8\times9.8}{3\times6.0^2\times1.00}}\text{ N}

F_{Cor} = \dfrac{2}{3}m\omega^2R\sqrt{5+\dfrac{8g}{3\omega^2R}} \approx 17\text{ N}

(5)

Check

Dimensionally, both boxed forces are (mass)(rad/s) $^2$ (m), a force, as required. Setting $\omega\to0$ collapses both to zero, as it must: a stationary sphere has no fictitious forces at all, in agreement with Step 2’s finding that the spin never affected the break-off point itself.

Remark

The whole problem splits cleanly because friction is absent: the real motion (Steps 2 to 3) is governed only by gravity and normal force and could not care less whether the sphere turns, while the two fictitious forces (Steps 4 to 5) are pure bookkeeping, computed after the fact from a motion that the spin never touched. A rough sphere would couple the two problems together from the very first line.