Circular and Rotational Motion

Sample questions

Showing 20 of 70 questions from this chapter. The full bank is in the chapter test below.

1. 1

   An arc of a circle of radius $r$ has length $S$ and subtends an angle $\theta$ at the centre. If $S = r$, then $\theta$ is

   • A.$\pi$ rad
   • B.$2\pi$ rad
   • C.✓ $1$ rad (correct answer)
   • D.$\pi/2$ rad

   Why: One radian is the angle subtended at the centre by an arc whose length equals the radius, so $\theta = S/r = r/r = 1$ rad. The value $\pi$ rad would require an arc of length $\pi r$, that is a semicircle.
2. 2

   The angular displacement of a rotating body is taken as positive when its sense of rotation is

   • A.✓ counter-clockwise, directed along the axis of rotation (correct answer)
   • B.clockwise, directed along the axis of rotation
   • C.counter-clockwise, directed along the tangent to the circle
   • D.counter-clockwise, directed radially outward

   Why: A small angular displacement is a vector whose direction lies along the axis of rotation, and it is given a positive sign when the rotation is counter-clockwise. The tangential and radial directions belong to linear velocity and acceleration, not to angular quantities.
3. 3

   The angular displacement of a body that makes one complete revolution is

   • A.$\pi$ rad
   • B.$2\pi$ rad or $180^\circ$
   • C.$\pi$ rad or $360^\circ$
   • D.✓ $2\pi$ rad or $360^\circ$ (correct answer)

   Why: In one revolution the point covers the whole circumference $2\pi r$, so $\theta = S/r = 2\pi$ rad, which is exactly $360^\circ$. The value $\pi$ rad is only half a revolution, that is $180^\circ$.
4. 4

   Angular velocity is defined as the rate of change of

   • A.angular acceleration with time
   • B.✓ angular displacement with time (correct answer)
   • C.linear displacement with time
   • D.the angle between two radii of a circle

   Why: Angular velocity is $\omega = \Delta\theta/\Delta t$, the rate at which angular displacement changes, and its SI unit is $\mathrm{rad\,s^{-1}}$. The rate of change of angular velocity is the angular acceleration, so the first option defines a different quantity.
5. 5

   The SI unit of angular acceleration is

   • A.✓ $\mathrm{rad\,s^{-2}}$ (correct answer)
   • B.$\mathrm{rad\,s^{-1}}$
   • C.$\mathrm{rad\,s}$
   • D.$\mathrm{m\,s^{-2}}$

   Why: Angular acceleration is the rate of change of angular velocity, so its unit is $(\mathrm{rad\,s^{-1}})/\mathrm{s} = \mathrm{rad\,s^{-2}}$. The unit $\mathrm{rad\,s^{-1}}$ belongs to angular velocity, and $\mathrm{m\,s^{-2}}$ to linear or tangential acceleration.
6. 6

   A point on a rigid body rotating about a fixed axis is at a distance $r$ from the axis. If the angular velocity of the body is $\omega$, the linear speed of the point is

   • A.$v = \frac{\omega}{r}$
   • B.$v = \frac{r}{\omega}$
   • C.$v = \omega^2 r$
   • D.✓ $v = r\omega$ (correct answer)

   Why: Dividing the arc length $S = r\theta$ by the time taken gives $v = r\omega$, so a point farther from the axis moves faster than a nearer one. The forms $\omega/r$ and $r/\omega$ invert the relation, and $\omega^2 r$ has the dimensions of acceleration, not speed.
7. 7

   A wheel of radius $0.40\ \mathrm{m}$ rotates with an angular acceleration of $3.0\ \mathrm{rad\,s^{-2}}$. The tangential acceleration of a point on its rim is

   • A.$7.5\ \mathrm{m\,s^{-2}}$
   • B.✓ $1.2\ \mathrm{m\,s^{-2}}$ (correct answer)
   • C.$0.12\ \mathrm{m\,s^{-2}}$
   • D.$3.0\ \mathrm{m\,s^{-2}}$

   Why: Since $v = r\omega$, the linear speed along the tangent changes at the rate $a_t = r\alpha = 0.40 \times 3.0 = 1.2\ \mathrm{m\,s^{-2}}$. Dividing $\alpha$ by $r$ instead of multiplying gives the reciprocal value $7.5\ \mathrm{m\,s^{-2}}$.
8. 8

   Two points on a rigid body rotating about a fixed axis lie at different distances from the axis. At any instant they have

   • A.the same linear speed and different angular speeds
   • B.the same linear speed and the same angular speed
   • C.✓ the same angular displacement and different linear speeds (correct answer)
   • D.the same linear acceleration and the same angular acceleration

   Why: Every point of a rigid body turns through the same angle in the same time, so all points share the same $\omega$ and $\alpha$; but $v = r\omega$ differs because the distances from the axis differ. Neither the linear speeds nor the tangential accelerations can be equal when $r$ is unequal.
9. 9

   A wheel starting from rest attains an angular velocity of $20\ \mathrm{rad\,s^{-1}}$ in $5.0\ \mathrm{s}$ with uniform angular acceleration. Its angular acceleration is

   • A.$100\ \mathrm{rad\,s^{-2}}$
   • B.✓ $4.0\ \mathrm{rad\,s^{-2}}$ (correct answer)
   • C.$2.5\ \mathrm{rad\,s^{-2}}$
   • D.$0.25\ \mathrm{rad\,s^{-2}}$

   Why: Using $\omega_f = \omega_i + \alpha t$ with $\omega_i = 0$ gives $\alpha = 20/5.0 = 4.0\ \mathrm{rad\,s^{-2}}$. Dividing the time by the angular velocity instead of the other way round gives the reciprocal $0.25\ \mathrm{rad\,s^{-2}}$, which is not an acceleration.
10. 10

    A body moves along a circular path at constant speed. Which statement about it is correct?

    • A.Its velocity is constant because its speed is constant
    • B.It is not accelerated because its speed is constant
    • C.It is accelerated outward along the radius
    • D.✓ It is accelerated inward along the radius (correct answer)

    Why: The direction of motion keeps changing, so the velocity vector changes and the body is accelerated even though its speed does not change; the acceleration points towards the centre of the circle. Velocity cannot be constant while its direction keeps changing, since velocity is a vector.
11. 11

    A grinding wheel rotating at $60\ \mathrm{rad\,s^{-1}}$ is brought to rest uniformly while turning through $30$ rad. Its angular retardation is

    • A.✓ $60\ \mathrm{rad\,s^{-2}}$ (correct answer)
    • B.$30\ \mathrm{rad\,s^{-2}}$
    • C.$120\ \mathrm{rad\,s^{-2}}$
    • D.$2.0\ \mathrm{rad\,s^{-2}}$

    Why: From $2\alpha\theta = \omega_f^2 - \omega_i^2$ we get $\alpha = (0 - 60^2)/(2 \times 30) = -60\ \mathrm{rad\,s^{-2}}$, so the retardation is $60\ \mathrm{rad\,s^{-2}}$. Omitting the factor 2 in front of $\theta$ gives the wrong value $30\ \mathrm{rad\,s^{-2}}$.
12. 12

    A wheel of radius $0.25\ \mathrm{m}$ turns through an angle of $4.0$ rad. The distance travelled by a point on its rim is

    • A.$0.0625\ \mathrm{m}$
    • B.$16\ \mathrm{m}$
    • C.✓ $1.0\ \mathrm{m}$ (correct answer)
    • D.$4.0\ \mathrm{m}$

    Why: The arc length is $S = r\theta = 0.25 \times 4.0 = 1.0\ \mathrm{m}$, so a point on the rim travels 1.0 m. The value $0.0625\ \mathrm{m}$ comes from dividing $r$ by $\theta$ instead of multiplying them.
13. 13

    The angular equations of motion $\omega_f = \omega_i + \alpha t$, $2\alpha\theta = \omega_f^2 - \omega_i^2$ and $\theta = \omega_i t + \frac{1}{2}\alpha t^2$ may be treated as scalar equations only when

    • A.the body rotates with constant angular velocity
    • B.✓ the axis of rotation is fixed, so all the angular vectors keep the same direction (correct answer)
    • C.the body is a uniform disc or cylinder
    • D.the angular acceleration is zero

    Why: These equations are the angular counterparts of the linear equations and they require all the angular vectors to lie along one fixed axis, so that they add like scalars. If the direction of the axis changes, the vector nature of $\omega$ and $\alpha$ can no longer be ignored.
14. 14

    The ratio of the angular speed of the minute hand of a watch to that of its hour hand is

    • A.✓ $12:1$ (correct answer)
    • B.$1:12$
    • C.$60:1$
    • D.$1:60$

    Why: The minute hand completes one revolution in 1 hour while the hour hand takes 12 hours, so $\omega_{min}/\omega_{hr} = (1/1)/(1/12) = 12$. The ratio $1:12$ is its inverse, which would make the hour hand the faster of the two.
15. 15

    An object in uniform circular motion completes $10$ revolutions in $2.0\ \mathrm{s}$. Its period and frequency are respectively

    • A.$2.0\ \mathrm{s}$ and $5.0\ \mathrm{Hz}$
    • B.$20\ \mathrm{s}$ and $0.20\ \mathrm{Hz}$
    • C.$0.50\ \mathrm{s}$ and $2.0\ \mathrm{Hz}$
    • D.✓ $0.20\ \mathrm{s}$ and $5.0\ \mathrm{Hz}$ (correct answer)

    Why: The period is the time for one revolution, $T = 2.0/10 = 0.20\ \mathrm{s}$, and the frequency is its reciprocal, $f = 10/2.0 = 5.0\ \mathrm{Hz}$. The first option mistakes the total time of 2.0 s for the period.
16. 16

    A constant force acts perpendicular to the velocity of a body moving with uniform speed. The force

    • A.changes both the speed and the direction of motion
    • B.changes the speed but not the direction of motion
    • C.✓ changes the direction of motion but not the speed (correct answer)
    • D.has no effect on the motion of the body

    Why: A force perpendicular to the velocity does no work on the body, so the speed stays the same while the direction changes continuously, and the path is bent into a circle. A force along the velocity is the one that changes the speed.
17. 17

    A ball tied to the end of a string is whirled in a horizontal circle. If the string snaps, the ball moves

    • A.✓ along the tangent to the circle at the point where the string broke (correct answer)
    • B.radially outward from the centre of the circle
    • C.along the circular path for a short time and then falls
    • D.towards the centre of the circle

    Why: With the tension gone there is no force to change the direction of the ball's velocity, so it continues in a straight line along the tangent at that instant. It does not fly radially outward; that impression comes from the inertia felt in the rotating frame.
18. 18

    A body of mass $m$ moves with uniform speed $v$ in a circle of radius $r$. Its centripetal acceleration is

    • A.$a_c = \frac{v}{r}$, directed outward
    • B.✓ $a_c = \frac{v^2}{r}$, directed towards the centre (correct answer)
    • C.$a_c = \frac{v^2}{r}$, directed along the tangent
    • D.$a_c = vr$, directed towards the centre

    Why: The centripetal acceleration has magnitude $v^2/r$ and always points towards the centre of the circle, which is why it changes only the direction of the velocity and not its magnitude. An acceleration along the tangent would be needed to change the speed.
19. 19

    The centripetal force acting on a body of mass $m$ moving in a circle of radius $r$ with angular velocity $\omega$ is

    • A.$F_c = m r^2 \omega$
    • B.$F_c = \frac{m\omega^2}{r}$
    • C.$F_c = \frac{m\omega}{r}$
    • D.✓ $F_c = m r \omega^2$ (correct answer)

    Why: Substituting $v = r\omega$ into $F_c = mv^2/r$ gives $F_c = mr\omega^2$. It is the first power of $r$ that appears, so a formula with $r$ in the denominator, or with $r^2$, cannot give the correct force.
20. 20

    A stone of mass $0.20\ \mathrm{kg}$ is whirled in a horizontal circle of radius $0.50\ \mathrm{m}$ at a constant speed of $4.0\ \mathrm{m\,s^{-1}}$. The tension in the string is

    • A.$0.80\ \mathrm{N}$
    • B.$3.2\ \mathrm{N}$
    • C.✓ $6.4\ \mathrm{N}$ (correct answer)
    • D.$1.6\ \mathrm{N}$

    Why: The tension provides the centripetal force, so $T = mv^2/r = (0.20 \times 16)/0.50 = 6.4\ \mathrm{N}$. The value $0.80\ \mathrm{N}$ comes from using $mv/r$ and forgetting to square the speed.