Circular Motion: 28 MCQs with answers

Radially inward toward the center

Centripetal acceleration always points toward the center of the circular path, causing the change in direction of velocity.

$a_c = \frac{v^2}{r}$

The standard formula for centripetal acceleration is $a_c = \frac{v^2}{r}$, where $v$ is the speed and $r$ is the radius.

m/s²

Acceleration is the rate of change of velocity, so its unit is meters per second squared (m/s²).

Perpendicular to the velocity vector and toward the center

Centripetal acceleration is perpendicular to the velocity (which is tangential) and points inward.

Changing the direction of the velocity

In uniform circular motion, speed is constant, so centripetal acceleration only changes the direction of velocity.

True

By definition, centripetal means 'center-seeking', so the acceleration points toward the center.

True

Since speed and radius are constant, $a_c = v^2/r$ is constant in magnitude.

The acceleration of an object moving in a circular path at constant speed, directed toward the center

Centripetal acceleration is defined as the acceleration experienced by an object moving in a circular path at constant speed, always directed toward the center of the circle.

$a_c = \frac{v^2}{r}$

The magnitude of centripetal acceleration is given by $a_c = \frac{v^2}{r}$, where $v$ is the speed and $r$ is the radius of the circular path.

A change in the direction of the velocity vector

In uniform circular motion, speed is constant, but the direction of velocity changes continuously. The change in velocity vector $\Delta \vec{v}$ arises from this change in direction.

$a_c = \omega^2 r$

Since $v = \omega r$, substituting into $a_c = \frac{v^2}{r}$ gives $a_c = \omega^2 r$.

Toward the center of the circle

Centripetal acceleration is always directed toward the center of the circular path, which is why it is called 'centripetal' (center-seeking).

They are similar triangles

The derivation uses the fact that the triangle formed by the initial and final velocity vectors is similar to the triangle formed by the radii, because both have the same angle between their corresponding sides.

The radius of the circular path

In the formula $a_c = \frac{v^2}{r}$, $r$ is the radius of the circular path.

$\frac{v^2}{r}$

The magnitude of centripetal acceleration is given by $a_c = \frac{v^2}{r}$.

m/s²

Acceleration is measured in meters per second squared (m/s²), so centripetal acceleration also has the SI unit m/s².

It is approximately equal to $v \Delta \theta$

For small angular displacements, the magnitude of the change in velocity is approximately $v \Delta \theta$, which leads to the derivation of $a_c = v \omega = \frac{v^2}{r}$.

The change in direction of the velocity

In uniform circular motion, speed is constant, so $\Delta \vec{v}$ arises solely from the change in direction of the velocity vector.

$T = \frac{2\pi r}{v}$

The circumference is $2\pi r$, so the time for one revolution is $T = \frac{2\pi r}{v}$.

Four times

Since $a_c \propto v^2$, doubling $v$ quadruples the acceleration.

$v \Delta \theta$

For small angles, $\Delta v \approx v \Delta \theta$ because the chord length approximates the arc length.

It is constant in magnitude but changes direction

The magnitude $v^2/r$ is constant, but the direction always points toward the center, so it rotates with the particle.

$a_c = \omega^2 r$

Using $v = \omega r$, we get $a_c = v^2/r = (\omega r)^2/r = \omega^2 r$.

8 m/s²

$a_c = v^2/r = (20)^2/50 = 400/50 = 8$ m/s².

It becomes double

Since $a_c \propto 1/r$, halving the radius doubles the acceleration.

Nine times

Since $a_c = \omega^2 r$, tripling $\omega$ multiplies $a_c$ by $3^2 = 9$.

The angular displacement $\Delta \theta$

The velocity vectors are perpendicular to the radii, so the angle between them equals the angle between the radii, which is $\Delta \theta$.

Both are isosceles with equal sides and the included angle is equal

The velocity vectors have equal magnitude $v$, and the radii have equal length $r$; the angle between the two velocity vectors equals the angle between the two radii, so the triangles are similar.