Rotational Dynamics Concepts Page - 6

Example

Torque about an axis for a rigid body

Example: A pulley of radius 0.4m$0.4m$ is being rotated by an external agent. A
crate of mass 100kg$100kg$ is dragged across a smooth horizontal ground by a light rope whose other end is being wrapped round the pulley. The pulley has a fixed smooth vertical axle and its moment of inertia about its axis is 24kgm2$24kg{m}^2$. The crate moves with an acceleration of 0.2m/s2$0.2\mathrm{m}/{\mathrm{s}}^2$

Solution:
Let τ0${\tau }_0$ (N-m) be the torque driving the pulley.
The linear acceleration of the crate is equal to the linear acceleration of the rope on the rim of the pulley. The angular acceleration of pulley is given by dividing  the linear acceleration a$a$ of the crate by the radius of the pulley. Thus-
α=ar=0.20.4=0.5rad/s2$\alpha ={\displaystyle {\displaystyle \frac{a}{r}}={\displaystyle \frac{0.2}{0.4}=0.5rad/s^2}}$
For crate: F$F$=$=$ma$ma$
Tension in rope T=$=$100(0.2)
or
T$T$=$=$20N$20N$
For rotation of pulley
Couple C$C$=$=$Iα$I\alpha$. Thus we have
τ0−(0.4)T=24(0.5)${\tau }_0-(0.4)T=24(0.5)$
or
τ0−(0.4)(20)=12${\tau }_0-(0.4)(20)=12$
or
τ0=20Nm${\tau }_0=20Nm$

Definition

Angular momentum in terms of angular velocity

Angular momentum is defined as L=Iω$L=I\omega$
where 
I:$I:$ Moment of inertia
ω:$\omega :$ Angular velocity
Note:
Similar to linear momentum (measure of the body to resist change in velocity), angular momentum is a measure of the body to resist change in angular velocity.

Example

Relation between rate of change on angular momentum with net torque

Example: A constant torque acting on a uniform circular wheel changes its angular momentum from L$L$ to 4L$4L$ in 4$4$ seconds. What is the torque acting on it?
Solution:As we know that, τ=ΔLΔt$\tau ={\displaystyle \frac{\mathrm{\Delta }L}{\mathrm{\Delta }t}}$
So, τ=4L−L4=3L4$\tau ={\displaystyle \frac{4L-L}{4}}={\displaystyle \frac{3L}{4}}$

Example

Angular Momentum

Example: A stone of mass m$m$ is projected with a velocity u$u$ at an angle of 45o${45}^o$ to the horizontal. What is its angular momentum about the point of projection when it is at its highest point?

Solution: Highest point B, velocity will only be in x$x$ direction
i.e. v⃗ =vcosθi^+0j^$\overrightarrow{v}=v\cos \theta \hat{i}+0\hat{j}$
while r⃗ =rxi^+ryj^$\overrightarrow{r}=r_x\hat{i}+r_y\hat{j}$
ry→=u2sin24502g,rx→=Range2=u2sin9002g${\displaystyle \overrightarrow{r_y}=\frac{u^2{\sin }^2{45}^0}{2g},\overrightarrow{r_x}=\frac{Range}{2}=\frac{u^2\sin {90}^0}{2g}}$
So, Angular momentum about point of projection when it is at highest point :
 A⃗ =m(r⃗ ×v⃗ )$\overrightarrow{A}=m(\overrightarrow{r}\times \overrightarrow{v})$
 A⃗ =m(rx→i^+ryj^)×(ucosθi^)$\overrightarrow{A}=m(\overrightarrow{r_x}\hat{i}+r_y\hat{j})\times (u\cos \theta \hat{i})$
 A⃗ =mryucosθ(−k^)=m(ucos450)(u24g)$\overrightarrow{A}=m{r}_{y}u\cos \theta (-\hat{k})=m(u\cos {45}^0)({\displaystyle \frac{u^2}{4g}})$
         =mu342√g.$={\displaystyle \frac{m{u}^3}{4\sqrt{2}g}}.$

Example

Angular Momentum of a particle with defined trajectory

Example: A particle performs uniform circular motion with an angular momentum L$L$. If the angular frequency f$f$ of the particle is doubled, and kinetic energy is halved, how it will affect its angular momentum?
Solution:L=mνR=mωR2$L=m\nu R=m\omega R^2$     -----(1)
now angular frequency is doubled
i.e.ω⇒2ω$\omega \Rightarrow 2\omega$
also KE is halved 12mω2R2→12(m2ω2R2)${\displaystyle \frac{1}{2}}m{\omega }^2{R}^2\to {\displaystyle \frac{1}{2}}\left({\displaystyle \frac{m}{2}}{\omega }^2{R}^2\right)$
ω1=ω,ω2=2ω${\omega }_1=\omega ,{\omega }_2=2\omega$
12[m2ω2R21]=[m2(2ω)2(R2)2]${\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{m}{2}}{\omega }^2{R}_1^2\right]=[{\displaystyle \frac{m}{2}}(2\omega {)}^2(R_2{)}^2]$
R212=4R22.${\displaystyle \frac{R_1^2}{2}}=4R_2^2.$
R2=R122√.$R_2={\displaystyle \frac{R_1}{2\sqrt{2}}}.$
put it in ---(1)
L′=m(2ω)(R22√)2$L^{\prime }=m(2\omega ){\left({\displaystyle \frac{R}{2\sqrt{2}}}\right)}^2$
=L4$={\displaystyle \frac{L}{4}}$

Example

Angular Momentum and its relation with Energy

Example: Two bodies with moment of inertia I1$I_1$ and I2$I_2$(I1>I2)$(I_1>I_2)$ have equal angular momentum. If the KE of rotation is E1$E_1$ and E2$E_2$ , what will be the relation between energies?

Solution: Krot=I2Iω2=I2I2ω2I$K_{rot}=\frac{I}{2}I{\omega }^2=\frac{I}{2}\frac{I^2{\omega }^2}{I}$  .....(1)

But, angular momentum L=Iω$L=I\omega$
∴$\therefore$ eqn(1) becomes Krot=L22I$K_{rot}=\frac{L^2}{2I}$
Since given L is constant, Krotα1I$K_{rot}\alpha \frac{1}{I}$
∴I1>I2⇒E2>E1$\therefore I_1>I_2\Rightarrow E_2>E_1$

Example

Angular momentum of a rigid body about an axis through its center of mass

Example:
A circular disc of mass 4kg$4kg$ and of radius 10cm$10cm$ is rotating about its natural axis at the rate of 5rad/sec$5rad/sec$, then find its angular momentum.Solution:
m = 4 kg
R = 10 cm
ω=5 rad/s$\omega =5rad/s$
L=Iω$L=I\omega$
=12M R2 ω$={\displaystyle \frac{1}{2}}M{R}^2\omega$
=12×4×(01)2×5$={\displaystyle \frac{1}{2}}\times 4\times (01{)}^2\times 5$
=0.01 kg m2/s$=0.01kg{m}^2/s$

Example

Angular momentum of a rigid body about an axis other than its center of mass

Example:
The length of seconds hand of watch is 1.5cm$1.5cm$ and its mass is 7×10−3g$7\times {10}^{-3}g$. Find its angular momentum.
Solution:

T=60s$T=60s$

ω=2π/60 rad/s$\omega =2\pi /60rad/s$

L=Iω=ml23ω$L=I\omega ={\displaystyle \frac{m{l}^2}{3}}\omega$

L=7×10−3×(1.5×10−2)23×2π/60$L={\displaystyle \frac{7\times {10}^{-3}\times (1.5\times {10}^{-2}{)}^2}{3}}\times 2\pi /60$

L=5.5×10−11kgm2/s

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