Circular Motion flash cards

1. 1.Define angular displacement and give its SI unit.

   Angular displacement $\theta$ is the angle swept by the radius vector of a particle moving on a circle. SI unit is the radian (rad), a dimensionless quantity. For a full revolution $\theta=2\pi$ rad.

   Hint: Angle turned, not distance moved.

2. 2.How is arc length $s$ related to angular displacement $\theta$?

   $s=r\theta$, where $r$ is the radius and $\theta$ is in radians. This linear-angular link underlies $v=r\omega$ and $a_t=r\alpha$.

   Hint: Only valid when $\theta$ is in radians.

3. 3.Is angular displacement a vector? What direction does it point?

   Small (infinitesimal) angular displacements behave as vectors; large ones do not (they don't commute). Direction is along the axis of rotation, given by the right-hand rule.

   Hint: Finite rotations fail vector addition.

4. 4.Define average and instantaneous angular velocity.

   Average $\bar\omega=\dfrac{\Delta\theta}{\Delta t}$; instantaneous $\omega=\dfrac{d\theta}{dt}$. SI unit rad/s. Direction along the rotation axis (right-hand rule).

   Hint: Rate of change of $\theta$.

5. 5.Relate linear speed $v$ to angular velocity $\omega$.

   $v=r\omega$. In vector form $\vec v=\vec\omega\times\vec r$. Larger radius means larger linear speed for the same $\omega$.

   Hint: Cross product gives direction tangent to circle.

6. 6.How is angular velocity related to time period $T$ and frequency $f$?

   $\omega=\dfrac{2\pi}{T}=2\pi f$. Here $T$ is the time for one revolution and $f=1/T$ is revolutions per second (Hz).

   Hint: One revolution = $2\pi$ rad.

7. 7.Define angular acceleration and give its unit.

   $\alpha=\dfrac{d\omega}{dt}=\dfrac{d^2\theta}{dt^2}$. SI unit rad/s$^2$. It is nonzero only when the rate of rotation changes (non-uniform circular motion).

   Hint: Rate of change of $\omega$.

8. 8.Relate tangential acceleration $a_t$ to angular acceleration $\alpha$.

   $a_t=r\alpha$. It is the component of acceleration along the tangent that changes the speed of the particle.

   Hint: Linear analogue of $\alpha$.

9. 9.Write the rotational kinematics equations for constant $\alpha$.

   $\omega=\omega_0+\alpha t$; $\theta=\omega_0 t+\tfrac12\alpha t^2$; $\omega^2=\omega_0^2+2\alpha\theta$. Direct analogues of linear $v=u+at$ etc.

   Hint: Replace $s,u,v,a$ by $\theta,\omega_0,\omega,\alpha$.

10. 10.What is centripetal acceleration and its formula?

    The acceleration directed toward the centre that continuously changes the velocity's direction: $a_c=\dfrac{v^2}{r}=\omega^2 r=v\omega$. Present in all circular motion.

    Hint: Points inward, magnitude $v^2/r$.

11. 11.Why does a particle in uniform circular motion still accelerate?

    Speed is constant but velocity's direction constantly changes. This change requires centripetal acceleration $a_c=v^2/r$ directed toward the centre, even though $|v|$ is fixed.

    Hint: Acceleration = change in velocity vector, not just speed.

12. 12.Distinguish centripetal and tangential acceleration by their roles.

    Centripetal $a_c=v^2/r$ is perpendicular to $v$ and changes its direction. Tangential $a_t=dv/dt$ is parallel to $v$ and changes its magnitude (speed).

    Hint: Perpendicular vs parallel to velocity.

13. 13.For non-uniform circular motion, what is the net (total) acceleration magnitude?

    $a=\sqrt{a_c^2+a_t^2}=\sqrt{\left(\dfrac{v^2}{r}\right)^2+\left(\dfrac{dv}{dt}\right)^2}$. The two components are mutually perpendicular.

    Hint: Vector sum of radial and tangential parts.

14. 14.What angle does the net acceleration make with the radius in non-uniform circular motion?

    The net acceleration makes angle $\phi$ with the radial (centripetal) direction where $\tan\phi=\dfrac{a_t}{a_c}$. It is not directed toward the centre unless $a_t=0$.

    Hint: Tangential over centripetal.

15. 15.Compare uniform and non-uniform circular motion.

    Uniform: constant speed, $a_t=0$, only $a_c$, $\alpha=0$. Non-uniform: speed changes, both $a_c$ and $a_t$ present, $\alpha\ne0$, net force has tangential component.

    Hint: Is the speed constant or not?

16. 16.State the direction of angular velocity and angular acceleration vectors for a speeding-up wheel.

    $\vec\omega$ lies along the axis (right-hand rule). If speeding up, $\vec\alpha$ is parallel to $\vec\omega$; if slowing down, $\vec\alpha$ is antiparallel to $\vec\omega$.

    Hint: Same axis; sign shows speeding/slowing.

17. 17.What is centripetal force and its magnitude?

    The net inward force required to keep a body of mass $m$ on a circular path: $F_c=\dfrac{mv^2}{r}=m\omega^2 r$. It is a requirement, provided by real forces (tension, gravity, friction, normal, etc.).

    Hint: Not a new force — it's a role played by real forces.

18. 18.Is centripetal force a separate fundamental force? Explain.

    No. It is the name for the net inward component of real forces (tension, friction, normal, gravity). It does no work on the particle since it is perpendicular to velocity in uniform circular motion.

    Hint: A role, not a new interaction.

19. 19.Does centripetal force do work in uniform circular motion? Why?

    No work. $F_c$ is perpendicular to displacement/velocity at every instant, so $W=\vec F\cdot\vec s=0$. Hence speed (and kinetic energy) stays constant.

    Hint: Force $\perp$ velocity.

20. 20.What provides the centripetal force for a car turning on a flat (unbanked) road?

    Static friction between tyres and road. Condition: $\dfrac{mv^2}{r}\le\mu_s mg$, giving maximum safe speed $v_{max}=\sqrt{\mu_s g r}$.

    Hint: Friction points toward centre.

21. 21.Derive the maximum speed on an unbanked road of radius $r$, friction $\mu_s$.

    Friction supplies centripetal force: $\mu_s mg\ge\dfrac{mv^2}{r}$, so $v_{max}=\sqrt{\mu_s g r}$. Independent of mass.

    Hint: Set friction = required centripetal force.

22. 22.Why are roads banked at curves?

    Banking lets the horizontal component of the normal force supply centripetal force, reducing reliance on friction. This allows safe turning at higher speeds and reduces tyre wear.

    Hint: Tilt the road so normal force leans inward.

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