Rotational Dynamics Concepts Page - 10

Definition

Relationship between angular acceleration and linear acceleration

Linear acceleration (a$a$) and angular acceleration (α$\alpha$) are related as:
a⃗ =α⃗ ×r⃗ $\overrightarrow{a}=\overrightarrow{\alpha }\times \overrightarrow{r}$

Example

Condition of slipping to determine nature of slipping

Example:  A sphere S$S$ rolls without slipping, moving with a constant speed on the plank P$P$. The friction between the upper surface of P$P$ and the sphere is sufficient to prevent slipping, while the lower surface of P$P$ is smooth and rests on the ground. Initially, P$P$ is fixed to the ground by a pin N$N$. What will happen if N$N$ is suddenly removed?

Solution: As the body is rolling without slipping at a constant speed, it is a state of pure rolling and there is no relative motion between the bottom most point of the sphere and the plank. Thus there will be no change in the motion of S$S$, and P$P$,they will still be at rest.

Example

Condition of slipping to determine direction of friction

Example: A sphere has to purely roll upwards. At an instant when the velocity of sphere is v$v$, what is the frictional force acting on it?

Solution:
Acceleration of a body rolling (pure)  down the inclined plane ,   a=gsinθ1+K$a={\displaystyle \frac{gsin\theta }{1+K}}$                            For  solid sphere     K=25$K={\displaystyle \frac{2}{5}}$
Let assume friction force to act upwards.Equation of motion,     
ma=mgsinθ−f$ma=mgsin\theta -f$    
 57mgsinθ=mgsinθ−f${\displaystyle \frac{5}{7}}mgsin\theta =mgsin\theta -f$⟹f=27mgsinθ$⟹f={\displaystyle \frac{2}{7}}mgsin\theta$ As f$f$ comes out to be positive thus our assumption is correct.

Example

No-slip condition where multiple points of contact are involved in pure rolling

Example: A spool of inner radius R$R$ and outer radius 3R$3R$ has a moment of inertia =MR2$=M{R}^2$ about an axis passing through its geometric center,
where M$M$ is the mass of the spool. A thread wound on the inner surface of the spool is pulled horizontally with a constant force =Mg$=Mg$. Find the acceleration of the point on the thread which is being pulled assuming that the spool rolls purely on the floor.
Solution:
We will equate the torques acting on the spool about the instantaneous center of rotation.
Mg(4R)=Iα$Mg(4R)=I\alpha$
I$I$ about the instantaneous center is MR2+M(3R)2=10MR2$M{R}^2+M(3R{)}^2=10M{R}^2$(use parallel axis theorem)
thus we get
4MgR=10MR2α$4MgR=10M{R}^2\alpha$
or
4g=10Rα$4g=10R\alpha$
or
4g=10Ra3R$4g=10R{\displaystyle \frac{a}{3R}}$
or
4g=103a$4g={\displaystyle \frac{10}{3}}a$
or
a=1210g$a={\displaystyle \frac{12}{10}}g$
This a$a$ is the acceleration of the center of mass.
Thus the acceleration of the thread at$a_t$ is given as the acceleration of point at a distance R$R$ from the center or 4R$4R$ from the instantaneous axis.
Thus we get at=4Rα=4Ra3R=43a$a_t=4R\alpha =4R{\displaystyle \frac{a}{3R}}={\displaystyle \frac{4}{3}}a$
or
at=431210g=1610g≈16m/s2${\displaystyle a_t=\frac{4}{3}\frac{12}{10}g=\frac{16}{10}g\approx 16m/s^2}$

Example

Conservation of energy equation for rigid bodies performing pure rolling where centre of mass moves in vertical circle

Example: A ball of 1$1$ gm$gm$ rolls down an inclined plane and describes a circle of radius 10$10$ cm$cm$ in the vertical plane on reaching the bottom. What is the minimum height of the inclined plane?

Solution: R=10cm.$R=10cm.$
Solid spherical ball.
for no slipping
V bottom =5gR−−−−√$=\sqrt{5gR}$
for compliting vertical circular
for finding this velocity apply Work Energy Theorem
mgH=12mV2+12×25×mR2(VR)2$mgH={\displaystyle \frac{1}{2}}m{V}^2+{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{2}{5}}\times m{R}^2({\displaystyle \frac{V}{R}}{)}^2$
10H=710V2$10H={\displaystyle \frac{7}{10}}{V}^2$
H=7100×5×10×10100$H={\displaystyle \frac{7}{100}}\times 5\times 10\times \frac{10}{100}$
H=0.35m$H=0.35m$
=35cm$=35cm$

Example

Force equations for rigid bodies performing pure rolling where centre of mass moves in a vertical circle

Example: When a bucket containing water is rotated fast in a vertical circle of radius R$R$, when the water in the bucket won't spill?

Solution:For water not to spill,    Mar=Mg$M{a}_r=Mg$
Mv2R=Mg⟹v=gR−−−√${\displaystyle \frac{M{v}^2}{R}}=Mg⟹v=\sqrt{gR}$
Now w=vR=gR−−√$w={\displaystyle \frac{v}{R}}=\sqrt{{\displaystyle \frac{g}{R}}}$w=2πν⟹v=g4π2R−−−−−√$w=2\pi \nu ⟹v=\sqrt{{\displaystyle \frac{g}{4{\pi }^{2}R}}}$ s−1$s^{-1}$

 (OR  r.p.s)v=g4π2R−−−−−√×60$v=\sqrt{{\displaystyle \frac{g}{4{\pi }^{2}R}}}\times 60$ r.p.m$r.p.m$
Thus  v=900gπ2R−−−−−√$v=\sqrt{{\displaystyle \frac{900g}{{\pi }^{2}R}}}$  r.p.m$r.p.m$

Example

Force and torque equations about different axes for pure rolling

Example: A solid cylinder rolls without slipping on an inclined plane inclined at an angle θ$\theta$. Find the linear acceleration of the cylinder. Mass of the cylinder is M$M$.

Solution:Force along the incline plane mgsinθ−f=ma$mgsin\theta -f=ma$ ....(1)
Writing torque about center fr=Iα$fr=I\alpha$  ...(2)
where I=mr22$I={\displaystyle \frac{m{r}^2}{2}}$ ...(3)
for pure rolling a=αr$a=\alpha r$ ...(4)
on solving the above equations we get a=2gsinθ3$a={\displaystyle \frac{2gsin\theta }{3}}$

Example

Condition for breaking off from surface in case of objects performing pure rolling on a curved surface

Example: A uniform ball of radius r$r$ rolls without slipping down from the top of a sphere of radius R$R$. Find the angular velocity of the ball at the moment it breaks off the sphere. The initial velocity of the ball is negligible.
Solution:Let us write the equation of motion for the center of the sphere at the moment of breaking-off.Since the normal reaction , N$N$, would be zero at moment of separation ,the force equation would be :
mv2(R+r)=mgcosθ${\displaystyle {\displaystyle \frac{m{v}^2}{(R+r)}}=mg\cos \theta }$,i.e. centripetal force would be provided by component of mg$mg$.
where, v$v$  is the velocity of the center of the sphere at that moment, and θ$\theta$ is the corresponding angle (figure shown below). The velocity v$v$ can be found from the energy conservation law, 
mgh=12mv2+12Iω2${\displaystyle mgh={\displaystyle \frac{1}{2}}m{v}^2+{\displaystyle \frac{1}{2}}I{\omega }^2}$
where, I$I$ is the moment of inertia of the sphere relative to the axis passing through the sphere's centre. i.e. I=25mr2${\displaystyle I=\frac{2}{5}m{r}^2}$.
In addition,
v=ωr$v=\omega r$; h=(R+r)(1−cosθ)$h=(R+r)(1-\cos \theta )$.
From the above four equations we obtain
ω=10g(R+r)17r2−−−−−−−−−−−−√

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