System of Particles and Rotational Motion

Rotational Kinematics, Dynamics and Rolling Motion

  • The angle swept by the radius vector is called Angular displacement
  • Radian is the unit of angular displacement
  • Line Joining the center to the position of particle in circular motion called radius vector.
  • If a particle completes ‘N’ circle its angular displacement θ = 2ΠN radians.
  • Relation between linear and angular displacement is S = rθ.
  • Equation of motion along circular path w = wo + αt
  • ω2 – ωo2 = 2αθ 
  • \tt \theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}
  • \tt \theta=\left(\frac{\omega_{0}+\omega}{2}\right)t
  • In a circular motion the force acting towards the centre called centripetal force
  • Centripetal force \tt F=\frac{mV^{2}}{r}=mr\omega^{2}
  • Centripetal acceleration \tt ac=\frac{V^{2}}{r}=r\omega^{2}
  • In non uniform circular motion then exists another acceleration called tangential acceleration at = rα.
  • Net acceleration \tt a=\sqrt{a_c^2+a_t^2}.
  • Tan θ =\tt \frac{at}{ac}
  • Net force \tt F= \sqrt{F_c^2+F_t^2}
  • Net force \tt F= \sqrt{\left(\frac{mv^{2}}{r}\right)^2+\left(mr\alpha\right)^2}
  • Maximum speed of a vehicle to take a turn on circular road \tt V= \sqrt{\mu gr}.
  • Maximum speed of vehicle \tt V= \sqrt{gr\ \tan\theta}.
  • Angle made by a conical pendulum with vertical \tt \tan\theta=\frac{r\omega^{2}}{g}.
  • Angular velocity of a conical pendulum \tt \omega=\sqrt{\frac{g\tan\theta}{r}}.
  • Time period of conical pendulum \tt t=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{r}{g\tan\theta}}.
  • Tension in the string of conical pendulum \tt T=m\sqrt{g^{2}+\left(\frac{v^{2}}{r}\right)^2}.
  • Velocity to any point in motion of a body in vertical circle \tt V=\sqrt{U^{2}-2gr\left(1-\cos\theta\right)} (u=initial velocity).
  • Tension in the string in vertical circle \tt T=mg \cos \theta+\frac{mv^{2}}{r}
  • Height of particle in vertical circle h = r (1 – cosθ)
  • Minimum velocity at bottom to complete vertical circle \tt U=\sqrt{5gr}.
  • Minimum velocity at top to complete vertical circle in critical condition \tt V=\sqrt{gr}.
  • Tension at the bottom of vertical circle in critical condition T = 6mg
  • Tension at the top of vertical circle in critical condition T = 0
  • Condition for leaving the circle in vertical circle \tt \sqrt{2gr}<u<\sqrt{5gr}.
  • The maximum height reached by particle in vertical circle h = u2 / 2g
  • The height reached by a particle where tension is zero. Q in vertical circular motion \tt h=\frac{u^{2}+rg}{3g}
  • Angle made by string with vertical where tension is zero \tt \cos\theta=\left(\frac{2}{3}-\frac{u^{2}}{3gr}\right).

View the Topic in this video From 0:59 To 55:00

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1. Equation of Rotational Motion
(i) ω = ω0 + αt
(ii) \theta = \omega_{0}t + \frac{1}{2} \alpha t^{2}
(iii) \omega^{2} = \omega_{0}^{2} + 2 \alpha \theta

2. Radius of gyration K = \sqrt{\frac{r_{1}^{2} + r_{2}^{2} + .... + r_{n}^{2}}{n}}

3. Body Inslipping posses only translatory kinetic energy K_{T} = \frac{1}{2}mv^{2}

4. In spinning, body possess only rotatory kinetic energy K_{R} = \frac{1}{2}I \omega^{2}

5. Rolling on an Inclined Plane
(i) Velocity at the lowest point: v = \sqrt{\frac{2gh}{1 + \frac{k^{2}}{R^{2}}}}
(ii) Acceleration in motion: a = \frac{g \sin \theta}{1 + \frac{k^{2}}{R^{2}}}
(iii) Time of descent : t = \frac{1}{\sin \theta}\sqrt{\frac{2h}{g}\left[1 + \frac{k^{2}}{R^{2}}\right]}