Question 82: Centre of Mass and Rotational Motion

Question

Two circular discs of radius each 10 cm are joined at their centres by a rod of length 30 cm and mass 600 gm as shown in figure.
If the mass of each disc is 600 gm and applied torque between two discs is $43 	imes 10^5$ dyne.cm, the angular acceleration of the discs about the given axis $A B$ is $\_\_\_\_$ $\mathrm{rad} / \mathrm{s}^2$.

Accepted Answer

To find the angular acceleration $(\alpha)$, we use the formula :
$$ 	au=\mathrm{I}_{	ext {total }} \cdot \alpha \Rightarrow \alpha=	au / \mathrm{I}_{	ext {total }} $$
where $\mathrm{I}_{	ext {total }}$ is the total moment of inertia of the assembly about the axis AB .
The rod has mass $\mathrm{M}=600 \mathrm{~g}$ and length $\mathrm{L}=30 \mathrm{~cm}$. The axis AB is perpendicular to the rod at a distance of 10 cm from one end.
The centre of mass (CM) of the rod is at 15 cm from either end.
The distance (d) between the axis AB and the CM of the rod is: $\mathrm{d}=15 \mathrm{~cm}-10 \mathrm{~cm}=5 \mathrm{~cm}$.
Using the Parallel Axis Theorem :
$$ I_{	ext {rod }}=\frac{M^2}{12}+{M d}^2 $$
$$\Rightarrow $$ $I_{\mathrm{rod}}=\frac{600 	imes 30^2}{12}+600 	imes 5^2$
$$\Rightarrow $$ $$ I_{\mathrm{rod}}=(50 	imes 900)+(600 	imes 25)=60,000 \mathrm{~g} \cdot \mathrm{~cm}^2 $$
The axis $A B$ is parallel to the diameter of the discs.
The moment of inertia of a disc about its diameter is $I_{	ext {dia }}= \frac{1}{4} \mathrm{MR}^2$.
For the left disc, distance from axis $A B$ is $r_1=10 \mathrm{~cm}$.
$$ I_1=\frac{1}{4} M R^2+M r_1^2=\frac{1}{4}(600)(10)^2+600(10)^2 $$
$$\Rightarrow $$ $$ \mathrm{I}_1=75000 \mathrm{~g} \cdot \mathrm{~cm}^2 $$
For the right disc, distance from axis AB is $\mathrm{r}_2=20 \mathrm{~cm}$.
$$ \mathrm{I}_2=\frac{1}{4} \mathrm{MR}^2+\mathrm{Mr}_2^2=\frac{1}{4}(600)(10)^2+600(20)^2 $$
$$ \mathrm{I}_2=255000 \mathrm{~g} \cdot \mathrm{~cm}^2 $$
So, total moment of inertia is,
$$ \mathrm{I}_{	ext {total }}=\mathrm{I}_{\mathrm{rod}}+\mathrm{I}_1+\mathrm{I}_2 $$
$$\Rightarrow $$ $I_{	ext {total }}=60000+75000+255000=390000 \mathrm{~g} \cdot \mathrm{~cm}^2$
$$\Rightarrow $$ $$ \mathrm{I}_{	ext {total }}=3.9 	imes 10^5 \mathrm{~g} \cdot \mathrm{~cm}^2 $$
Given Torque $	au=43 	imes 10^5$ dyne $\cdot \mathrm{cm}$.
$$ \alpha=\frac{	au}{\mathrm{I}_{	ext {total }}}=\frac{43 	imes 10^5}{3.9 	imes 10^5} $$
$$\Rightarrow $$ $$ \alpha=\approx 11.025 \mathrm{rad} / \mathrm{s}^2 $$
The nearest integer value is 11 . Hence, the correct option is (D).

Answer

100

Answer

22

Answer

27

Answer

11

Centre of Mass and Rotational Motion

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