Rotational Dynamics Concepts Page - 9

Result

Comparison of quantities in rotational and translational motion

Linear Motion	Rotational Motion
Displacement x$x$	Angular Displacement θ$\theta$
Velocity v=dxdt$v={\displaystyle \frac{dx}{dt}}$	Angular velocity ω=dθdt$\omega ={\displaystyle \frac{d\theta }{dt}}$
Acceleration a=dvdt$a={\displaystyle \frac{dv}{dt}}$	Angular acceleration α=dωdt$\alpha ={\displaystyle \frac{d\omega }{dt}}$
Mass M	Moment of inertia I$I$
Force F=Ma$F=Ma$	Torque τ=Iα$\tau =I\alpha$
Work dW=Fds$dW=Fds$	Work dW=τdθ$dW=\tau d\theta$
Kinetic Energy K=12mv2$K={\displaystyle \frac{1}{2}}m{v}^2$	Kinetic energy K=12Iω2$K={\displaystyle \frac{1}{2}}I{\omega }^2$
Power P=Fv$P=Fv$	Power P=τω$P=\tau \omega$
Linear momentum p=Mv$p=Mv$	Angular momentum L=Iω$L=I\omega$

Example

Problem on rotation plus translation where more than one force contributes to torque

Example: Inner and outer radii of a spool are r$r$ and R$R$ respectively. A
thread is wound over its inner surface and spool is placed over a rough horizontal surface. Thread is pulled by a force F$F$ as shown in figure. In case of pure rolling, how will be the motion of spool with winding/unwinding of thread?
Solution: Since, the spool rolls over the horizontal surface, instantaneous axis
of rotation passes through the point of contact of spool with the horizontal surface. About the instantaneous axis of rotation, moment produced by F is clockwise. Therefore, the spool rotates clockwise. In that case, acceleration will be rightward and thread will wind. If rotational motion of the spool is considered about its own, then the resultant moment on it must be clockwise. But moment produced by the force is anticlockwise and its magnitude is equal to F1/r$F_1/r$. Hence, moment produced by the friction (about its own axis) must be clockwise
and its magnitude must be greater than Fr$Fr$. It is possible only when friction acts leftwards.

Definition

Instantaneous axis of rotation

The instantaneous axis of rotation is the axis fixed to a body undergoing planar movement that has zero velocity at a particular instant of time. At this instant, the velocity vectors of the trajectories of other points in the body generate a circular field around this axis which is identical to what is generated by a pure rotation.

For a body performing pure rotation instantaneous axis of rotation lies at the point of contact as given in the above figure.

Example

Instantaneous axis of rotation to find angular velocity

Example: A cylinder is rolling without sliding over two horizontal planks (surfaces) 1 and 2. If the velocities of the surfaces A$A$ and B$B$ are −vi^$-v\hat{i}$ and 2vi^$2v\hat{i}$ respectively. Then what will be the position of instantaneous axis of rotation and also what will be the angular velocity?

Solution:Let v1$v_1$ be the velocity of COM of cylinder and w$w$ be its angular velocity.
Let C be the position of instantaneous axis of rotation.
2v=w(CB)⟹2v=(2R−x)w$2v=w(CB)⟹2v=(2R-x)w$           ...........(a)
v=xw$v=xw$       ............(b)
2(wx)=(2R−x)w⟹x=2R3$2(wx)=(2R-x)w⟹x={\displaystyle \frac{2R}{3}}$
Also there is no slipping at A and B.
Thus  2v=v1+Rw$2v=v_1+Rw$    and  v=Rw−v1$v=Rw-v_1$Eliminating v1$v_1$,  
we get  w=3v2R$w={\displaystyle \frac{3v}{2R}}$

Definition

Pure rolling and its conditions

Rolling without slipping is a combination of translation and rotation where the point of contact is instantaneously at rest. When an object experiences pure translational motion, all of its points move with the same velocity as the center of mass; that is in the same direction and with the same speed.

For a ball of radius r$r$ moving with translational velocity v$v$ and rotating with angular velocity w$w$ condition for pure rolling is:
v=wr$v=wr$

Example

Use of formula of kinetic energy of a body in pure rotation

Example: Consider a uniform disc of mass 4 kg performing pure rolling with velocity 5 m/s on a fixed rough surface. Comment on the kinetic energy of the upper half of the disc.

Solution: T.K.E. of disc
=12mv2+12Iω2${\displaystyle =\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2}$
=12mv2+12×mr22×(vr)2=+34mv2=75J${\displaystyle =\frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{m{r}^2}{2}\times {\left(\frac{v}{r}\right)}^2=+\frac{3}{4}m{v}^2=75J}$
Velocity of particles of upper half is more than that of lower half
hence kinetic energy of upper half will be more than 38mv2=37.5J$\frac{3}{8}m{v}^2=37.5J$

Example

Problem on Rolling Friction

Example: A boy is pushing a ring of mass 2$2$ kg and radius 0.5$0.5$ m$m$ with a stick as shown in the figure. The stick applies a force of 2$2$ N$N$ on the ring and rolls it without slipping with an acceleration of 0.3$0.3$ m/s2$m/s^2$ The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is P10${\displaystyle \frac{P}{10}}$.  What is the value of P$P$ ?

Solution:
The FBD is shown in the figure.
Moment of inertia I=mR2=2×(0.5)2=0.5kg m2$I=m{R}^2=2\times (0.5{)}^2=0.5kg{m}^2$
N−fs=ma=2×0.3=0.6$N-f_s=ma=2\times 0.3=0.6$
2−0.6=fs$2-0.6=f_s$
1.4=fs$1.4=f_s$
a=Rα=RτI=R(fs−fk)RI=(0.5)2(fs−fk)0.5$a=R\alpha =R{\displaystyle \frac{\tau }{I}}={\displaystyle \frac{R(f_s-f_k)R}{I}}={\displaystyle \frac{(0.5{)}^2(f_s-f_k)}{0.5}}$
fs−fk=0.3×0.5(0.5)2=0.6$f_s-f_k={\displaystyle \frac{0.3\times 0.5}{(0.5{)}^2}}=0.6$
1.4−fk=0.6$1.4-f_k=0.6$
⇒fk=0.8$\Rightarrow f_k=0.8$
fk=μN=P10×2$f_k=\mu N={\displaystyle \frac{P}{10}}\times 2$
⇒0.8=P10×2$\Rightarrow 0.8={\displaystyle \frac{P}{10}}\times 2$
⇒P=4$\Rightarrow P=4$

Example

Conditions on coefficient of friction to facilitate pure rolling

Example: A bowling ball of uniform density is projected along a horizontal with a velocity v0$v_0$ so that it initially slides without rolling. The ball has mass m$m$ and coefficient of static friction μd${\mu }_d$ with the floor. Ignore air-friction. Let t$t$ be the time at which the ball begins to roll without sliding and v$v$ be the velocity of the ball when this happens.

Solution:Friction not only causes v to decrease but produces a torque which gives an angular acceleration α$\alpha$ causing ω$\omega$ to increase.
Ffr=μN=μmg$F_{fr}=\mu N=\mu mg$
τ=FfrR=Iα$\tau =F_{fr}R=I\alpha$
⇒μmgR=25mR2α$\Rightarrow \mu mgR={\displaystyle \frac{2}{5}}m{R}^2\alpha$
⇒α=5μg2R$\Rightarrow \alpha ={\displaystyle \frac{5\mu g}{2R}}$
Hence, from equation of translational motion, v=u+at$v=u+at$ we have  v=v0+at=v0−μgt$v=v_0+at=v_0-\mu gt$ as u=v0$u=v_0$ and a=−μg$a=-\mu g$
From equation of rotational motion, ω=ω0+αt=0+5μg2Rt=5μg2Rt$\omega ={\omega }_0+\alpha t=0+{\displaystyle \frac{5\mu g}{2R}}t={\displaystyle \frac{5\mu g}{2R}}t$ as ω0=0${\omega }_0=0$
In case of rolling without sliding, v=ωR$v=\omega R$
⇒(v0−μgt)=R(5μg2R)t$\Rightarrow (v_0-\mu gt)=R({\displaystyle \frac{5\mu g}{2R}})t$
v0=μgt+52μgt=72μgt$v_0=\mu gt+{\displaystyle \frac{5}{2}}\mu gt={\displaystyle \frac{7}{2}}\mu gt$
t=2v07μg$t={\displaystyle \frac{2v_0}{7\mu g}}$
Substituting the value of t in v=v0−μgt$v=v_0-\mu gt$
we get v=v0−μg2v07μg=57v0

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