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Revision:Motion Of A Rigid Body

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Motion of a Rigid Body

These notes are based on the requirement of the M5 A Level mathematics module.


Contents

Rotation of a rigid body

\displaystyle K.E._{\mathsf{of\ a\ body}} = \frac{1}{2}I\omega^2

\displaystyle P.E._{\mathsf{of\ a\ body}} = \M g h_{\mathsf{of\ centre\ of\ mass}}


You should know that conservation of energy should be used with such things.

Moment of the resultant force about an axis:

\displaystyle L = I_{\mathsf{about\ the\ axis}} . \theta'' (proof to that is in Heinemann M5 book, page 85, if you don’t have the book, and still need the proof, please contact me). \theta '' is the angular acceleration.


The components of acceleration of centre of mass (G) which is rotating about axis through O are:

  • \displaystyle r \theta '^2 along GO
  • \displaystyle r \theta '' perpendicular to OG

To find the force that the body is doing on the axis: put force X at the axis in the direction GO, and force Y at O perpendicular to OG. Then resolve any other forces on the body in the directions OY and OX, then use:

  • radial forceresultant \displaystyle = m r \theta '^2
  • transverse forceresultant \displaystyle = m r \theta ''

hence find X and Y, and then find their resultant force (if required to find the magnitude of the force acting on the axis).


Angular momentum

Angular momentum = moment of momentum \displaystyle = I \theta '

(How to derive that is in the same book page 94, also contact me if needed.)


As angular momentum = moment of momentum, then angular momentum for a body moving in a straight line is: its momentum × the distance from the axis.


Conservation of angular momentum

Same as conservation of momentum, angular momentumbefore = angular momentumafter


Example

A rod ( mass m, length 4a ) is free to rotate about a vertical axis through its centre, it rests on a smooth horizontal table, a particle of mass 3m moving in a straight line perpendicular to the rod with speed u ms[sup]-1[/sup] hits the rod at distance a from one of its ends. The particle sticks to the rod. Find the angular speed of the body after collision.


Initial angular momentumrod = 0

Initial angular momentumparticle \displaystyle = (3m.u).a


Final angular momentumrod \displaystyle = I_{\mathsf{rod}} \omega

Final angular momentumparticle \displaystyle = I_{\mathsf{particle}} \omega

\displaystyle I_{\mathsf{rod}} = \frac{1}{3}.m.(2a)^2 = \frac{4}{3}ma^2

\displaystyle I_{\mathsf{particle}} = m a^2


Initial Angular momentumtotal = Final angular momentumtotal

\displaystyle 3mua = \frac{4}{3} ma^2 \omega + ma^2 \omega

\displaystyle 3u = \frac{7}{3} a \omega

\displaystyle \omega = \frac{9u}{7a}


Effect of an Impulse on a rigid body that is free to rotate about an axis

You know that:

\displaystyle L = I \theta ''


Integrate with respect to t:

\displaystyle \int_{t_1}^{t_2} L\ dt = \int_{\omega _1}^{\omega _2} I \theta ''\ dt


\displaystyle \int_{t_1}^{t_2} L\ dt = [I \theta’]_{\omega_1}^{\omega_2}


\displaystyle \int_{t_1}^{t_2} L\ dt = I \Delta \omega

usually \displaystyle L = F . r so:

\displaystyle \int_{t_1}^{t_2} F . r\ dt = I \Delta \omega

\displaystyle F . r (\Delta t) = I \Delta \omega

\displaystyle F.(\Delta t) = \frac{I \Delta \omega r}

\displaystyle \mathsf{Impulse} = \frac{I\Delta \omega}{r}


Pendulums

Simple pendulum

\displaystyle \mathsf{Period} = 2\pi \sqrt{\frac{l}{g}}


Compound pendulum

\displaystyle \mathsf{Period} = 2\pi \sqrt{\frac{I}{mgh}}

where h is the distance between axis and C.M. and I is moments of inertia about the axis.


Or it might be written in another form:

\displaystyle \mathsf{Period} = 2\pi \sqrt{\frac{k^2 + h^2}{gh}}

where k is the radius of gyration.


To find the simple pendulum with the same period of a certain compound pendulum: use T = T,

i.e. \displaystyle 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{I}{mgh}} which is simplified to \displaystyle l = \frac{I}{mh}.


Or

\displaystyle 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{k^2 + h^2}{gh}} which is simplified to \displaystyle l = \frac{k^2 + h^2}{h}.


Comments

Originally written by yazan_l on TSR forums.

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