Irodov 1.53: Three Rim Points, One Rigid Body – Amit & Deepali Sinha Foundation

Problem Irodov 1.53

A ball of radius $R=10.0$ cm rolls without slipping down an inclined plane, starting from rest, its center’s speed increasing at a constant rate $w=2.50$ cm/s $^2$ . At the instant $t=2.00$ s after release, consider the ball’s contact point with the incline, $O$ ; the point $A$ diametrically opposite $O$ on the rim; and the point $B$ on the rim at the end of the diameter perpendicular to $OA$ , in the sense that carries the ball’s motion. Find:

the velocities of $A$ , $B$ and $O$ at that instant;
the accelerations of $A$ , $B$ and $O$ at that instant.

The ball at t=2.00 s in the local frame of the incline: the contact O, the center and the point A on one line perpendicular to the rolling direction, B at the end of the perpendicular diameter, w the center's constant acceleration. — The ball at $t=2.00$ s in the local frame of the incline: the contact $O$ , the center and the point $A$ on one line perpendicular to the rolling direction, $B$ at the end of the perpendicular diameter, $\vect w$ the center’s constant acceleration.

Given

$R=10.0$ cm
$w=2.50$ cm/s $^2$
$t=2.00$ s
starts from rest

Find

(a)

\vect v_A,\vect v_B,\vect v_O

, (b)

\vect w_A,\vect w_B,\vect w_O

Solution

Step 1The center’s speed and the ball’s angular velocity at

t

The center starts from rest with constant acceleration $w$ , and rolling without slipping ties the angular velocity to the center’s speed through $\Omega=v_C/R$ .

v_C(t) = wt, \qquad \Omega(t) = \frac{v_C(t)}{R} = \frac{wt}{R}

v_C = (2.50)(2.00) = 5.0\ \text{cm/s}, \qquad \Omega = \frac{5.0}{10.0} = 0.50\ \text{rad/s}

Step 2Velocity of a rim point, and the three named points

A clockwise angular velocity is $\vect\Omega=-\Omega\vect k$ in these axes. Writing a rim point’s position relative to the center in components and expanding the cross product with the standard unit vector products,

\begin{align*}\vect v(x,y) &= \vect v_C+\vect\Omega\times\vect r \\ &= v_C\vect i+(-\Omega\vect k)\times(x\vect i+y\vect j) \\ &= v_C\vect i+\Omega y\,\vect i-\Omega x\,\vect j\end{align*}

Collecting the two components gives the velocity field of the rolling ball, which Step 4 needs again.

\vect v(x,y) = (v_C+\Omega y)\,\vect i - \Omega x\,\vect j

(1)

The three points sit at these positions relative to the center, with $\Omega R=v_C$ by the rolling condition:

\vect r_A = R\vect j, \qquad \vect r_B = R\vect i, \qquad \vect r_O = -R\vect j, \qquad \Omega R = v_C. \tag{2}

Substituting each in turn,

\begin{align*}\vect v_A &= (v_C+\Omega R)\vect i = 2v_C\,\vect i \\ \vect v_B &= v_C\,\vect i-\Omega R\,\vect j = v_C(\vect i-\vect j) \\ \vect v_O &= (v_C-\Omega R)\vect i = \vect 0\end{align*}

\begin{gathered}v_A = 2v_C = 10.0\ \text{cm/s}\\ v_B = \sqrt{2}\,v_C = 7.1\ \text{cm/s}\\ v_O = 0\end{gathered}

(3)

Here $\vect v_A$ points along the rolling direction and $\vect v_B$ at $45^\circ$ to it.

Step 3The ball’s angular acceleration

Differentiating $\Omega(t)=wt/R$ gives a constant angular acceleration, clockwise like $\Omega$ itself.

\beta = \frac{d\Omega}{dt} = \frac{w}{R} = \frac{2.50}{10.0} = 0.25\ \text{rad/s}^2

Step 4Acceleration of a rim point, and the three named points

Differentiating (1) amounts to the rigid body acceleration formula: the center’s own acceleration, plus a tangential term $\vect\beta\times\vect r$ from the spin-up, plus the centripetal term $-\Omega^2\vect r$ . The angular acceleration is clockwise like $\Omega$ , so $\vect\beta=-\beta\vect k$ , and the cross product expands exactly as it did in Step 2.

\begin{align*}\vect w(x,y) &= \vect w_C+\vect\beta\times\vect r-\Omega^2\vect r \\ &= w\vect i+(-\beta\vect k)\times(x\vect i+y\vect j)-\Omega^2(x\vect i+y\vect j) \\ &= w\vect i+\beta y\,\vect i-\beta x\,\vect j-\Omega^2 x\,\vect i-\Omega^2 y\,\vect j\end{align*}

Collecting the components gives the acceleration field.

\vect w(x,y) = \bigl(w+\beta y-\Omega^2 x\bigr)\vect i - \bigl(\beta x+\Omega^2 y\bigr)\vect j

(4)

Substituting the three positions (2) into (4), with $\beta R=w$ from Step 3,

\begin{align*}\vect w_A &= (w+\beta R)\vect i - \Omega^2 R\,\vect j = 2w\,\vect i-\Omega^2R\,\vect j \\ \vect w_B &= (w-\Omega^2 R)\vect i - \beta R\,\vect j = (w-\Omega^2R)\vect i-w\,\vect j \\ \vect w_O &= (w-\beta R)\vect i + \Omega^2 R\,\vect j = \Omega^2R\,\vect j\end{align*}

Evaluating the recurring combinations,

\Omega^2R = (0.50)^2(10.0) = 2.5\ \text{cm/s}^2, \qquad \beta R = w = 2.5\ \text{cm/s}^2, \qquad w-\Omega^2R = 0,

so the $\vect i$ component of $\vect w_B$ vanishes, leaving only its $-\beta R\vect j=-w\vect j$ term: the two numbers agree at this particular instant, but $\vect w_B$ ’s surviving component is the tangential $\beta R$ term, not the centripetal $\Omega^2R$ one.

\begin{gathered}w_A = \sqrt{(2w)^2+(\Omega^2R)^2} = 5.6\ \text{cm/s}^2\\ w_B = w = 2.5\ \text{cm/s}^2\\ w_O = \Omega^2R = 2.5\ \text{cm/s}^2\end{gathered}

(5)

Both of the last two are normal to the incline: $\vect w_B$ into the surface, $\vect w_O$ away from the surface, toward the center.

Check

The classic rolling identities hold: the point diametrically opposite the contact moves at exactly twice the center’s speed, $v_A=2v_C=10.0$ cm/s, and the contact itself has zero velocity, $v_O=0$ , which is the rolling constraint restated. Dimensionally every acceleration above is a sum of a $\beta R$ or $w$ term (cm/s $^2$ ) with an $\Omega^2R$ term ((rad/s) $^2\cdot$ cm = cm/s $^2$ ), consistent throughout.

Remark

Point $O$ has zero velocity yet a nonzero acceleration, $\Omega^2R=2.5$ cm/s $^2$ directed toward the center, and this is no coincidence of the numbers: the tangential term $\beta R$ always exactly cancels the center’s own $w$ in $\vect w_O$ , because rolling ties $\beta=w/R$ for any constant $w$ . So the contact of a uniformly accelerating rolling body always accelerates purely centripetally, $\Omega^2R$ toward the center, no matter how hard the body is being sped up.