System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 56

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Question 1
1 / -0

A circular platform is mounted on a vertical frictionless axle. Its radius is $$r = 2 { m }$$ and its moment of inertia is $$I=200{ kg }-{ m }^{ { 2 } }.$$ It is initially at rest. A $$70 { kg }$$ man stands on the edge of the platform and begins to walk along the edge at speed $$v _ { 0 } = 1.0 { m } / { s }$$ relative to the ground. The angular velocity of the platform is

A
$$1.2\quad { rad }/{ s }$$

B
$$0.4\quad { rad }/{ s }$$

C
$$2.0\quad { rad }/{ s }$$

D
$$0.7\quad { rad }/{ s }$$

Solution
Question 2
1 / -0

A thin wire of length L and uniform linear density $$\rho$$ used to form a ring. The M.I. of this ring about tangential axis laying in its planes is:

A
$$\frac { \rho L ^ { 3 } } { 8 \pi ^ { 2 } }$$

B
$$\frac { \rho L ^ { 3 } } { 16 \pi ^ { 2 } }$$

C
$$\frac { 3 \rho L ^ { 3 } } { 12 \pi ^ { 2 } }$$

D
$$\frac { 3 \rho L ^ { 3 } } { 8 \pi ^ { 2 } }$$

Solution

REF.Image
Mass of ring = $$\rho \times L$$
Then,
MOI about $$OO'= (\rho L)(\frac{L}{2\pi })^{2}$$
$$=\frac{\rho L^{3}}{4\pi ^{2}}$$ $$2\pi r=L$$, density = $$\rho $$ $$r=\frac{L}{2\pi }$$
Then,
MOI about $$NN'=\frac{\rho L^{3}}{4\pi ^{2}}\times \frac{1}{2}=\frac{\rho L^{3}}{8\pi ^{2}}$$
(By perpendicular axis theorem)
Finally by parallel axis theorem
MOI about $$MM'=\frac{\rho L^{3}}{8\pi ^{2}}+(\rho L)(\frac{L}{2\pi })^{2}$$
$$=\frac{\rho L^{3}}{8\pi ^{2}}+\frac{\rho L^{3}}{4\pi ^{2}}=\frac{\rho L^{3}+2\rho L^{3}}{8\pi ^{2}}=\frac{3\rho L^{3}}{8\pi ^{2}}$$
Question 3
1 / -0

If the radius of solid sphere is $$\frac{1}{2}\sqrt{\frac{5}{7}}$$ m, Thin its radius of gyration when the axis is along the tangent is -

A
$$\frac{5}{14}$$ m

B
$$\frac{1}{2}$$ m

C
$$\frac{5}{7}$$ m

D
1 m

Solution

$$\begin{array}{l} \text { Given, radius of solid sphere }=\frac{1}{2}\sqrt{\frac{5}{7}} \text { m. }(r) \text { .[ }\left.\operatorname{mass}=m\right] \\\text { we know, moment of inertia about diameter }=\frac{2}{5} m r^{2} \text { . }\end{array}$$

$$\begin{array}{l}\text { By parallel axis theorem: } \\\text { moment of inertia about tangent }(I)=\frac{2}{5} mr^{2}+m r^{2}=\frac{7}{5} mr^{2}\end{array}$$

$$\begin{array}{l}\text { Let, radius of gyration is } k \text { . }\\\text { Then, we can write, } I=m k^{2} \Rightarrow k=\sqrt{\frac{I}{m}} \\\text { So, } k=\sqrt{\frac{\frac{7}{5} m r^{2}}{m}}=\sqrt{\frac{7}{5}} r=\frac{1}{2} m\end{array}$$
Question 4
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The momentum of body is numerically equal to its kinetic energy, The velocity of the body is :

A
1 m/s

B
2 m/s

C
4 m/s

D
8 m/s

Solution
Question 5
1 / -0

Two disc one of density $$7.2{ g/cm }^{ 3 }$$ and the other of density $$8.9{ g/cm }^{ 3 }$$ are of same mass and thickness. Their moments of inertia are in the ratio:-

A
$$\dfrac { 8.9 }{ 7.2 } $$

B
$$\dfrac { 7.2 }{ 8.9 } $$

C
$$\left( 8.9\times 7.2 \right) :1$$

D
$$1:\left( 9.9\times 7.2 \right) $$

Solution

$$\begin{array}{l}\text { Let, these two discs be } A \text { and } B\text { , such that } \\\text { density of } A\left(\rho_{A}\right)=7\cdot 2 g m / \mathrm{cm}^{3} \\\text { and, density of } B\left(\rho_{B}\right)=8.9\mathrm{gm}/\mathrm{cm}^{3}\end{array}$$

$$\begin{array}{l}\text { Given } \rightarrow \quad m_{A} \text { and } m_{B} \text { are equal and their thickness are equal. } \\\text { Let, radius of } \operatorname{disc} A \text { } r \text { }_A \text{and that of B} \text { is } r_{B} \text { . } \\\text { Also,} \text { let thickness of each discs be t. }\end{array}$$

$$\begin{aligned}m_{A}=m_{B} &\Rightarrow\rho_{A}\times\left(volume\right)_{A}=\rho_{B} \times(\text { volume })_{B} \\& \Rightarrow 7 \cdot 2 \times\left[\pm x \pi r_{A}^{2}\right]=8 \cdot 9 \times\left[ +_{}x\pi r_{B}^{2}\right]\end{aligned}$$

$$\begin{aligned}\therefore \quad \frac{r_A^{2}}{r_{B}^{2}} &=\frac{8\cdot 9}{7 \cdot 2} \\\text { Thul, ratio of their moment of inertia }=\frac{m_{A} r{A}^{2}}{m_{B} r_{B^{2}}} &=\frac{r_{A}^{2}}{r_{B}^{2}} \quad\left(\because m_{A}=m_{B}\right) \\&=\frac{8 \cdot 9}{7 \cdot 2}\end{aligned}$$
Question 6
1 / -0

Two identical particle move towards each other with velocity $$2v$$ and $$v$$ respectively. The velocity of centre of mass is

A
$$v$$

B
$$v/3$$

C
$$v/2$$

D
Zero

Solution
Question 7
1 / -0

A 5kg body collides with another stationary body. after the collision , the bodies moves in the same direction with one-third of the velocity of the first body . the mass of the second body will be-

A
$$5 kg$$

B
$$10 kg$$

C
$$15 kg$$

D
$$20 kg$$

Solution
Question 8
1 / -0

Find I of the rod along the given axis. Suppose the mass of the rod is $$m$$.

A
$$ \dfrac{ mL^{2} }{ 12 } $$

B
$$ \dfrac{ mL^{2} }{ 16 } $$

C
$$ \dfrac{ mL^{2} }{ 20 } $$

D
$$ \dfrac{ mL^{2} }{ 24 } $$
Question 9
1 / -0

From a disc of radius R, a concentric circular portion of radius r is cut out so as leave an annular disc of mass M. The moment of inertia of this annular disc about the axis perpendicular to its plane and passing through its center of gravity is

A
$$1/2 M(R^{2}+r^{2})$$

B
$$1/2 M(R^{2}-r^{2})$$

C
$$1/2 M(R^{4}+r^{4})$$

D
$$1/2 M(R^{4}-r^{4})$$

Solution

$$\begin{array}{l}\text { Given, mass of left annular disc }=M .\\\qquad \text { so, }\text { density }=\frac{M}{\pi\left(R^{2}-r^{2}\right)}\end{array}$$

$$\begin{array}{l}\text { Now, moment of inertia }\left(I_{1}\right)\text { of disc of radius } R \\\qquad \text { about central axis }=\frac{m_{1} R^{2}}{2} \quad\left(m_{1} \rightarrow \text { its mass }\right) \\\text { and, moment of inertia }\left(I_{2}\right) \text { of removed disc } \\\text { about the central axis }=\frac{m_{2} r^{2}}{2}\left(m_{2} \rightarrow\right.\text { its mass) }\end{array}$$

$$\begin{aligned}\therefore \text { moment of inertia of remaluing portion } &=I_{1}-I_{2} \\\text { (I) } &=\frac{1}{2}\left[M_{1} R^{2}-M_{2} r^{2}\right]\end{aligned}$$

$$\begin{array}{l}\text { Now, } M_{1}=\pi R^{2} \times \frac{M}{\pi \left(R^{2}-r^{2}\right)}=M\left(\frac{R^{2}}{R^{2}-r^{2}}\right) \\\text { and, } m_{2}=\pi r^{2} \times \frac{M}{\pi \left(R^{2}-r^{2}\right)}=M\left(\frac{r^{2}}{R^{2}-r^{2}}\right)\end{array}$$

$$\begin{aligned}\text { Thus, } I=\frac{1}{2}\left[\frac{M R^{4}}{\left(R^{2}-r^{2}\right)}-\frac{M r^{4}}{\left(R^{2}r^{2}\right)}\right] &=\frac{M}{2\left(R^{2}-r^{2}\right)}\left(R^{4}-r^{4}\right) \\&=\frac{1}{2} M\left(R^{2}+r^{2}\right)\end{aligned}$$
Question 10
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A uniform rod of mass 6M and length 6I is bent to make an equilateral hexagon. If M.I. about an axis passing through the centre of mass and perpendicular to the plane of hexagon is

A
$$5mI^2$$

B
$$6mI^2$$

C
$$4mI^2$$

D
$$mI^2/12$$

Solution

$$\begin{array}{l}\text { Clearly, length and mass of each edge }=\ell\text { and } M . \\\text { Consider portion BC : } \\\text { moment of inertia of BC about axis passing its } \\\text { centre and perperdicular to its plane is } \\\qquad \frac{m l^{2}}{12} \text { . }\end{array}$$

$$\begin{array}{l}\text { Thus, moment of inertia about given axis of } B C=\frac{m l^{2}}{12}+m d^{2} \\\text { (I } \left._{\text {BC }}\right) \quad \text { (Parallel axis theo. )} \\\therefore I_{B C}=\frac{m l^{2}}{12}+m\left(\frac{\sqrt{3} l}{2}\right)^{2}=\frac{m l^{2}}{12}+\frac{3 m l^{2}}{4}=\frac{5}{6} m l^{2}\end{array}$$

$$\text { Hence, total moment of inertia of system }=6 I_{B C}=5 m l^{2}$$