Question bank — Relation to linear quantities - v = rω, a_t = rα, a_c = rω²

True — $v=r\omega$ is linear in $\omega$, so doubling $\omega$ with $r$ fixed exactly doubles $v$.

False — $a_c=r{\omega }^2$ depends on $\omega$ squared, so doubling $\omega$ multiplies $a_c$ by four, not two.

True — angular velocity describes the whole body's turning; every point sweeps the same angle per second regardless of $r$.

False — $v=r\omega$ grows with $r$, so the rim point outruns an inner point even though they share one $\omega$.

False — velocity is a vector; its direction keeps turning, so $a_c=r{\omega }^2\ne 0$ even when $v$ never changes.

True — $a_t=r\alpha$ measures change in speed; constant speed means $\alpha =0$, so $a_t=0$ while $a_c$ still lives.

False — ${\vec{a}}_t$ points along the track and ${\vec{a}}_c$ points inward, so they meet at $90^{\circ }$. Two arrows at a right angle form the legs of a right triangle whose diagonal is the true total, so they combine by Pythagoras: $a=\sqrt{a_t^2+a_c^2}$ (see the master figure). Adding them as plain numbers would wrongly assume they point the same way.

True — slowing means speed is decreasing, so ${\vec{a}}_t$ (which changes speed) points against the motion arrow $\vec{v}$; $\alpha$ is then negative relative to $\omega$.

True — $a_c=r{\omega }^2$ scales with $r$ at shared $\omega$, so the ratio is $0.6/0.2=3$, exactly matching the ratio of their speeds $v$.

Wrong: $\omega$ is identical for every point on a rigid body. The faster rim speed comes entirely from the larger $r$ in $v=r\omega$, not a larger $\omega$.

Wrong: it is $a_c=r{\omega }^2$. Check units: $r^2\omega$ gives ${\text{m}}^2/\text{s}$, but acceleration must be ${\text{m/s}}^2$, which $r{\omega }^2=\text{m}\cdot {\text{s}}^{-2}$ delivers.

Wrong at fixed $\omega$: rewrite as $a_c=r{\omega }^2$, which increases with $r$. The $v^2/r$ form hides that $v$ itself grows with $r$, so $a_c\propto r$ for a rigid body.

Wrong: the formula is built for radians. Converting to degrees breaks it; if given degrees, convert to radians first.

Wrong: $a_t$ has no $\omega$ in it at all. It depends only on $r$ and $\alpha$; a fast but steadily-spinning wheel ($\alpha =0$) has $a_t=0$.

Wrong: only $a_c$ points inward. When the body speeds up or slows ($a_t\ne 0$), the total tilts off the radius by the tangential part.

Wrong: constant $\omega$ still means $a_c=r{\omega }^2\ne 0$, requiring an inward (centripetal) net force to keep the point on its circle.

Because we differentiate the point's distance-along-the-path and pull $r$ out of the derivative; a rigid body keeps $r$ fixed, so no extra $dr/dt$ term appears.

$a_t$ handles speed changes and $a_c$ handles direction changes; a point on a circle is always turning, so its velocity direction is always changing — $a_c$ can never vanish while it moves.

Because centripetal acceleration is born from the velocity's direction turning; only vector differentiation captures that change, which a scalar speed alone would miss.

The radian is defined so that arc length equals radius times angle (see Radian measure and arc length), which makes $v=r\omega$ and $a_t=r\alpha$ come out with no conversion factors — degrees would inject clumsy $\pi /180$ terms.

$a_t=r\alpha$: the angular acceleration is shared by the whole body, but multiplying by a larger $r$ stretches it into a larger tangential acceleration.

Picture the master figure but with $\omega$ climbing: the inward arrow ${\vec{a}}_c=r{\omega }^2$ stretches with the square of the spin, while the tangential arrow ${\vec{a}}_t=r\alpha$ doesn't feel $\omega$ at all. So the inward arrow quickly grows enormous next to the fixed-length tangential one, and the total acceleration swings almost fully toward the center — geometrically the diagonal collapses onto the inward leg. This is why fast-spinning rims are pulled apart from the center rather than dragged along.

Newton's laws and kinetic energy live in linear language ($v$, $a$); the radius $r$ is the dictionary that converts the body-wide angular quantities into the point-by-point linear ones.

All three are zero, since each is $r$ times an angular quantity — the point on the axis doesn't move at all, no matter how fast the body spins.

Only $a_c=r{\omega }^2$ survives; $a_t=r\alpha =0$. The motion is uniform — steady speed, constant inward pull.

$a_c=r{\omega }^2=0$ at that instant, but $a_t=r\alpha \ne 0$; the point is momentarily not turning direction yet is gaining speed.

Yes — $v=r\omega$ is a product, so a large radius can compensate for a small angular velocity, giving a substantial tangential speed.

No — $a_c=r{\omega }^2$ depends on ${\omega }^2$, which is positive either way; centripetal acceleration always points inward regardless of spin direction.

Using $a_c=v^2/r$, as $r\to \infty$ the centripetal acceleration $\to 0$ — the path straightens out and behaves like uniform straight-line motion.

Yes, when $r\alpha =r{\omega }^2$, i.e. $\alpha ={\omega }^2$; there the total acceleration points exactly $45^{\circ }$ between the tangent and the inward radius (right-hand panel of the master figure).