Rotational Dynamics Concepts Page - 8

Example

Conservation of Angular Momentum of more than one rigid body

Example: The above set of figures show components of a mechanical system in
(a) and the assembled system in (b). There are two balls of mass m and 2m, a massless tube of length L$L$, a spring of natural length l(>L)$l(>L)$ and an inextensible, massless rope of length L$L$. In the assembly the string holds the two balls in position, keeping the spring compressed. At a time t=t0$t=t_0$, the string is suddenly broken and the balls are released to move. Mark the incorrect statement. Consider the momentum and angular momentum after and before string breaks.

Solution:In the table frame momentum of individual balls is conserved. As there was no external force on the system , Angular momentum is conserved in table reference frame as well as centre of mass frame. When we look the system from the centre of mass frame the system would look like a mass body with string rotating. Therefore, In the centre of mass frame ,the angular momentum of each of the mass is seperately conserved. As there is no external force even linear momentum should conserve.Therefore total linear momentum of system is conserved in table reference frame. But we cannot say in the table reference frame, momentum of individual balls is conserved or not.

Example

Conservation of Angular Momentum incorporating relative motion between bodies

Example: A man of mass m1$m_1$ stands on the edge of a horizontal uniform disc of mass m2$m_2$ and radius R$R$ which is capable of rotating freely about a stationary vertical axis passing through its centre. At a certain moment the man starts moving along the edge of the disc; he shifts over an angle φ′${\phi }^{\mathrm{\prime }}$ relative to the disc and then stops. In the process of motion the velocity of the man varies with time as ω′(t)${\omega }^{\mathrm{\prime }}(t)$. Assuming the dimensions of the man to be negligible, find the angle through which the disc had turned by the moment the man stopped.
Solution:Since there is no torque acting on the disc-man system, angular momentum is conserved. Let angular velocity of disc be w(t)$w(t)$ and that of the man w.r.t the disc be wo(t)$w_o(t)$
By conservation of angular momentum:
m1ω′(t)R2+m2R22w(t)=0$m_1{\omega }^{{}^{\prime }}(t)R^2+{\displaystyle \frac{m_2{R}^2}{2}}w(t)=0$
m1R2(wo(t)+w(t))+m2R22w(t)=0$m_1{R}^2(w_o(t)+w(t))+{\displaystyle \frac{m_2{R}^2}{2}}w(t)=0$
w(t)=−2m12m1+m2wo(t)$w(t)=-{\displaystyle \frac{2m_1}{2m_1+m_2}}{w}_o(t)$
∫w(t)dt=−2m12m1+m2∫wo(t)dt$\int w(t)dt=-{\displaystyle \frac{2m_1}{2m_1+m_2}}\int w_o(t)dt$
φ=−2m12m1+m2φ′$\phi =-{\displaystyle \frac{2m_1}{2m_1+m_2}}{\phi }^{{}^{\prime }}$

Example

Torque equations in reference frame of axis

Example: The handle of a door is at a distance 40cm$40cm$ from axis of rotation.
If a force 5N$5N$ is applied on the handle in a direction 300${30}^0$  with plane of door, then what is the value of torque?
Solution:We know that the torque is given using the relation
τ=FR sin θ$\tau =FRsin\theta$
Or
τ=(5N)(0.4m)(sin30o)=1Nm$\tau =(5N)(0.4m)(sin{30}^o)=1Nm$

Example

Work Done by torque

Example: What will be the work done in rotating a body from angle θ1${\theta }_1$ to angle θ2${\theta }_2$ by a constant torque τ$\tau$ ?

Solution: Work done in angular displacement is given as
∫θ2θ1dW=∫θ2θ1τdθ${\displaystyle {\int }_{{\theta }_1}^{{\theta }_2}dW={\int }_{{\theta }_1}^{{\theta }_2}\tau d\theta }$. For constant terms we get the relation as W=τ(θ2−θ1)$W=\tau ({\theta }_2-{\theta }_1)$

Example

Power delivered by torque

Example: An electric motor rotates a wheel at a constant angular velocity ω$\omega$ while opposing torque is t$t$. What is the power of that electric motor?

solution: Power of Motor P⃗ =τ⃗ ×ω⃗ $\overrightarrow{P}=\overrightarrow{\tau }\times \overrightarrow{\omega }$
The motor is rotating at a constant angular velocity ω$\omega$ against a Torque t.
∴P=t×ω=tω$\therefore P=t\times \omega =t\omega$

Example

Fixed axis rotation with constant torque for rigid bodies

Example: A rigid body is rotating about a vertical axis. In t$t$ second, the
axis gradually becomes horizontal. But the rigid body continues to make v$v$ rotations per second throughout the time interval of 1s$1s$. If the moment of inertia I$I$ of the body about the axis of rotation can be taken as constant, then what is the torque acting on the body?

Solution: τ⃗ =Iα⃗ $\overrightarrow{\tau }=I\overrightarrow{\alpha }$ or
 τ⃗ =I(ω2→−ω1→)t$\overrightarrow{\tau }={\displaystyle \frac{I(\overrightarrow{{\omega }_2}-\overrightarrow{{\omega }_1})}{t}}$
But |ω2−ω1→|=ω22+ω21−−−−−−√=2√ω=2√×2πv$|{\omega }_2-\overrightarrow{{\omega }_1}|=\sqrt{{\omega }_2^2+{\omega }_1^2}=\sqrt{2}\omega =\sqrt{2}\times 2\pi v$
τ=It×22√πv$\tau ={\displaystyle \frac{I}{t}}\times 2\sqrt{2}\pi v$ or τ=22√πvIt$\tau ={\displaystyle \frac{2\sqrt{2}\pi vI}{t}}$

Example

Kinetic Energy in rotation plus translation

Example: What is the ratio of rotational kinetic energy and translatory kinetic energy of a rolling circular disc?

Solution: rotational kinetic energy=12Iω2=12(12MR2)ω2${\displaystyle \frac{1}{2}}I{\omega }^2={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1}{2}}M{R}^2\right){\omega }^2$
Translational K.E=12mν2${\displaystyle \frac{1}{2}}m{\nu }^2$
V=Rω$V=R\omega$
⇒ratio=12(12mν2)12mν2=12$\Rightarrow ratio={\displaystyle \frac{{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1}{2}}m{\nu }^2\right)}{{\displaystyle \frac{1}{2}}m{\nu }^2}}={\displaystyle \frac{1}{2}}$

Definition

Total velocity and acceleration of a point in rigid body in rotation plus translation

A body in combined translational rotational motion, velocity (or acceleration) of all points are a vectors sum of velocity (or acceleration) of center of mass and velocity (or acceleration) due to rotation about the center of mass.
v⃗ =v⃗ cm+r⃗ ×ω$\overrightarrow{v}={\overrightarrow{v}}_{cm}+\overrightarrow{r}\times \omega$
a⃗ =a⃗ cm+r⃗ ×α$\overrightarrow{a}={\overrightarrow{a}}_{cm}+\overrightarrow{r}\times \alpha$
Example:
A sphere is rolling without slipping on a fixed horizontal plane surface with a linear speed of vo$v_o$. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point.
Then,
vo=Rω$v_o=R\omega$vC=vo+Rω=2vo$v_C=v_o+R\omega =2v_o$
vB=vo+0=vo$v_B=v_o+0=v_o$
vA=vo−Rω=0

BookMarks

Page 1  Page 2  Page 3  Page 4  Page 5  Page 6  Page 7  Page 8  Page 9  Page 10
Page 11  Page 12