Question

1) Two discs of moments of intertia $I_{1}and{I}_2$ about their respective axes (normal to the disc and passing throug the centre), and rotating with angular speed ${\omega }_{1}and{\omega }_2$ are brought into contact face to face with their axes of rotion coincident

(A) Dose the law of conservation of angular momentum apply to the situation ? why ?

Hint : ''Torque by external force about the axis of rqatation''

Formula used : $\overrightarrow{\tau }=\frac{\overrightarrow{dL}}{dt}$


2) Two discs of moments of intertia $I_{1}and{I}_2$ about their respective axes (normal to the disc and passing throug the centre), and rotating with angular speed ${\omega }_{1}and{\omega }_2$ are brought into contact face to face with their axes of rotion coincident

(B) Find the angular speed of the two-disc system.

Hint : ''Conservation of angular momentum''

Formula used : $L=I\omega$

3) Two discs of moments of intertia $I_{1}and{I}_2$ about their respective axes (normal to the disc and passing throug the centre), and rotating with angular speed ${\omega }_{1}and{\omega }_2$ are brought into contact face to face with their axes of rotion coincident

(C) Calculate the loss in kinetic energy of the system in the process.

Hint : $\Delta K=K_f-K_i$

Formula used : $K=\frac{1}{2}I{\omega }^2$

4) Two discs of moments of intertia $I_{1}and{I}_2$ about their respective axes (normal to the disc and passing throug the centre), and rotating with angular speed ${\omega }_{1}and{\omega }_2$ are brought into contact face to face with their axes of rotion coincident

(D) Account for loss in energy.

Hint : ''Loss in kinetic energy''

Formula used : $K=\frac{1}{2}I{\omega }^2$

Answer 1

A) Yes, because there is no net external torque on the system. External force, gravitational, acts parallel to the axis of rotation, and friction is internal , hence, produce no change in angular momentum.


B) There is no net external torque on the system. External force, gravitational, acts parallel to the axis of rotation, and friction is internal , hence, produce no change in angular momentum.

So angular momentum is conserved about axis of rotation

$L_i=L_f$

$I_1{\omega }_1+I_2{\omega }_2=I\omega$

$I_1{\omega }_1+I_2{\omega }_2=(I_1+I_2)\omega$

$\omega =\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}$

Final Answer : $\omega =\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}$

C) There is loss of kinetic energy as heat, as discs are brought in contact with each other.

$K_i=\frac{1}{2}{I}_1{\omega }_1^2$+$\frac{1}{2}{I}_2{\omega }_2^2$

As they are brought in contact together,

So angular momentum is conserved about axis of rotation

$L_i=L_f$

$I_1{\omega }_1+I_2{\omega }_2=I\omega$

$I_1{\omega }_1+I_2{\omega }_2=(I_1+I_2)\omega$

$\omega =\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}$

$K_f=\frac{1}{2}(I_1+I_2){\omega }^2$

$K_f=\frac{1}{2}(I_1+I_2){\left(\frac{I_1{\omega }_1+I_2{\omega }_2}{I_1+I_2}\right)}^2$

$K_f=\frac{1}{2}\frac{(I_1{\omega }_1+I_2{\omega }_2{)}^2}{(I_1+I_2)}$

$\Delta K=K_f-K_i$

$=\frac{1}{2}\frac{(I_1{\omega }_1+I_2{\omega }_2{)}^2}{(I_1+I_2)}-\frac{1}{2}(I_1{\omega }_1^2+I_2{\omega }_2^2)$

$=-\frac{I_1{I}_2}{2(I_1+I_2)}({\omega }_1-{\omega }_2{)}^2$

Final Answer :

$K_f=\frac{1}{2}(I_1+I_2)\frac{(I_1{\omega }_1+I_2{\omega }_2{)}^2}{(I_1+I_2{)}^2}=\frac{1}{2}\frac{(I_1{\omega }_1+I_2{\omega }_2{)}^2}{I_1+I_2}$

$K_i=\frac{1}{2}(I_1{\omega }_1^2+I_2{\omega }_2^2)$


$\Delta K=K_f-K_i=-\frac{I_1{I}_2}{2(I_1+I_2)}({\omega }_1-{\omega }_2{)}^2$


D) The loss in kinetic energy is due to the work against the friction between the two discs.