Rotation of a Rigid Body under Constant Acceleration

  • Draw, analyze and interpret angular displacement-time, angular velocity-time and angular acceleration-time graphs.

  • Solve simple problems involving uniform angular motion and uniformly accelerated motion by using appropriate equations.

\[\begin{aligned} \omega &= \omega_0 + \alpha t \\ \theta &= \theta_0 +\omega_0 t + \dfrac{1}{2} \alpha_0 t^2 \\ \omega^2 &= \omega_0^2 + 2\alpha_0 (\theta - \theta_0) \end{aligned}\]

where, \[\begin{aligned} \omega_0 &= \mathrm{Initial\ angular\ velocity}\\ \omega &= \mathrm{Final\ angular\ velocity}\\ t &= \mathrm{time}\\ \theta_0 &= \mathrm{Initial\ angular\ displacement}\\ \theta &= \mathrm{angular\ displacement}\\ \alpha &= \mathrm{angular\ acceleration}\\ \end{aligned}\]

Solved Examples

Solved Example:

33-2-01

A bike has a front wheel of radius 0.25 m. It starts from rest, and accelerates down a hill with linear acceleration a= 2.0 m/s$^2$. At t= 1.0 sec, what is the angular velocity $\omega$ of the bike wheel?

Solution:
\[\alpha =a/r = \dfrac{2}{0.25} = 8.0\ rad/s^2\] \[\omega = \omega_0 + \alpha t =0 + (8)(1) = 8 rad/s\]

Correct Answer: C

Solved Example:

33-2-02

A cylinder (r = 0.14 m, I$_{cm}$ = 2.4 $\times$ 10$^{-2}$ kg.m$^2$, M = 1.5 kg) starts at rest and rolls without slipping down a plane with an inclination angle of $\theta$ =15$^\circ$. Find the time it takes to travel 1.4 m.

Solution:

33-2-02

Use Newton's second law, \[\sum F_{x}=mg\sin\theta-F_{FR}=ma\] Torque about center of mass \[\sum \tau=rF_{FR}=I\alpha=\dfrac {Ia}{r}\] \[F_{FR}=\dfrac {Ia}{r^{2}}\] \[mg\sin\theta-I\dfrac {a}{r^{2}}=ma\] \[a=\dfrac {r^{2}mg\sin\theta}{I+mr^{2}}=1.4m/s^{2}\] Now using, \begin{align*} x&=x_{0}+v_{0}t+\dfrac {1}{2}at^{2}\\ 1.4&=0+0+\dfrac {1}{2}\left( 1.4\right) t^{2}\\ t&=1.4s \end{align*}

Correct Answer: B