System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 57

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Question 1
1 / -0

A uniform cube of side a and mass m rests on a rough horizontal surface. A horizontal force F is applied normal to face at a point that is directly above the centre of the face at a height $$\cfrac{\alpha}{4}$$ above the centre. The minimum value of F for which the cube begins to topple above an edge without sliding is:

A
$$\cfrac {1}{4}mg$$

B
$$2mg$$

C
$$\cfrac{1}{2}mg$$

D
$$\cfrac{2}{3}mg$$

Solution
Question 2
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A solid cylinder of mass $$M$$ and radius $$R$$ is rolling without slipping on a horizontal plane with a speed $$v$$. it the rolls up an inclined plane of inclination $$\theta$$ to a maximum height given by

A
$$\frac { v ^ { 2 } } { 2 g }$$

B
$$\frac { v ^ { 2 } } { 2 g } \sin \theta$$

C
$$\frac { 3 v ^ { 2 } } { 4 g }$$

D
$$\frac { 3 v ^ { 2 } } { 4 g } \sin \theta$$

Solution

$$\begin{array}{r}\text { Given - } * \text { Solid cylinder mass }=M \text { and } \\\text { radius } R \text { rolling without slip. }\end{array}$$

$$\begin{array}{l}\Rightarrow \text { kE kinetic energy }=\text { Translation } k E \\ +\text { Rotational } k E \\\Rightarrow k E_{T}=\frac{1}{2} M v^{2}+\frac{1}{2} I w^{2} \quad I \rightarrow \text { Moment of Inertia } \\\qquad k E+=\frac{1}{2} M v^{2}+\frac{1}{2}\left(\frac{M R^{2}}{2}\right) w^{2}=\frac{3}{4} M v^{2}\quad(v=\text { tuw })\end{array} $$
$$\begin{array}{l}\text { For maximum Height by energy conservation } \\\qquad M g H=\frac{3}{4} M v^{2} \quad \Rightarrow H=\frac{3 v^{2}}{4 g}\end{array}$$
Question 3
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The velocity of centre of mass of the system as shown in the figure.

A
$${(\dfrac{2-2\sqrt{3}}{3})}\hat{i}-\dfrac{1}{3} \hat{j}$$

B
$${(\dfrac{2+2\sqrt{3}}{3})}\hat{i}-\dfrac{2}{3} \hat{j}$$

C
$$4\hat{i}$$

D
None of these

Solution

Let $$V_x$$ be the $$x-$$ component of velocity of CM, $$V_x$$

$$=\dfrac{m_1\times v_{1x}+m_2\times v_{2x}}{m_1+m_2}=\dfrac{1\times 2+2\times 2\times \dfrac{\sqrt 3}{2}}{1+2}=\dfrac{2+2\sqrt 3}{3}\,m/s$$

Let $$V_y$$ be the $$y-$$ component of velocity of CM, $$V_y$$

$$=\dfrac{m_1\times v_{1y}+m_2\times v_{2y}}{m_1+m_2}=\dfrac{1\times 0+2\times 2\times \dfrac 12}{1+2}=\dfrac 23\,m/s$$

Velocity of CM

$$\sqrt{V_x^2-V_y^2}=(\dfrac{2+2\sqrt 3}{2})\hat i-\dfrac 23 \hat j$$
Question 4
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The end A of a ladder AP of length 5 m, kept inclined to a vertical wall is slipping over a horizontal surface, when A is at a distance of 3 m from the wall. If velocity of centre of mass at this moment is $$1.25 m/s$$ then velocity of the lowest point A of the ladder at tis instant is :-

A
1.25 m/s

B
0 m/s

C
1 m/s

D
2 m/s

Solution
Question 5
1 / -0

A system of binary stars of masses $$m_{A}$$ and $$m_{B}$$ are moving in circular orbits of radii $$r_{A}$$ and $$r_{B}$$ respectively. If $$T_{A}$$ and $$T_{B}$$ are the time periods of masses $$m_{A}$$ and $$m_{B}$$ respectively, then

A
$$\frac{T_{A}}{T_{B}}=(\frac{r_{A}}{r_{B}})^{3/2}$$

B
$$T_{A} > T_{B} (if r_{A} > r_{B})$$

C
$$T_{A} > T_{B} (if m_{A} > m_{B})$$

D
$$T_{A} = T_{B} $$
Question 6
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if a force$$ 10\hat { i } +15\hat { j } +25\hat { K }$$ acts on a system and gives an acceleration $$2\hat { i } +3\hat { j } -5\hat { K } $$ to the centre of mass of the system, the mass of the system is :

A
5 units

B
$$\sqrt { 38 } units$$

C
$$5\sqrt { 38 } units $$

D
None

Solution

Given,
$${ F }_{ cm }=10\hat { i } +15\hat { j } +25\hat { k } \\ { a }_{ cm }=2\hat { i } +3\hat { j } -5\hat { k } $$
We know that force is multiplication of mass and acceleration therefore,
$${ F }_{ cm }=m\hat { a } \\ 10\hat { i } +15\hat { j } +25\hat { k } =m(2\hat { i } +3\hat { j } -5\hat { k } )\\ m=\frac { 10\hat { i } +15\hat { j } +25\hat { k } }{ 2\hat { i } +3\hat { j } -5\hat { k } } \\ m=\frac { \sqrt { { 10 }^{ 2 }+{ 15 }^{ 2 }+{ 25 }^{ 2 } } }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 } } } \\ m=\frac { \sqrt { 950 } }{ \sqrt { 38 } } \\ m=\frac { 5\sqrt { 38 } }{ \sqrt { 38 } } =5\quad units$$
Hence, correct option is $$(A)$$
Question 7
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Three man $$ A, B \& C $$ of mass $$ 40 \mathrm{kg}, 50 \mathrm{kg} \& 60 \mathrm{kg} $$ are standing on a plank of mass $$90 \mathrm{kg} $$ .which is kept on a smooth horizontal plane. If $$ \mathrm{A} \& \mathrm{C} $$ exchange their positions then mass $$ \mathrm{B} $$ will shift

A
$$1 / 3 \mathrm{m} $$ towards left

B
$$1 / 3 m $$ towards right

C
will not move w.r.t. ground

D
$$7 / 3 \mathrm{m} $$ towards left
Question 8
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A rod of moment of inertia I and length L is suspended from a fixed end and given small oscillations about the point of suspension , the restoring torque is found to be -(mgL/2) $$sin\theta $$ What will be the angular equation of motion of the SHM

A
$$\dfrac { { d }^{ 2 }\theta }{ { dt }^{ 2 } } +\left( mgL/I \right) \theta =0$$

B
$$\dfrac { { d }^{ 2 }\theta }{ { dt }^{ 2 } } +\left( mgL/2I \right) \theta =0$$

C
$$\dfrac { { d }^{ 2 }\theta }{ { dt }^{ 2 } } +\left( 2mgL/I \right) \theta =0$$

D
$$\dfrac { { d }^{ 2 }\theta }{ { dt }^{ 2 } } +\left( mgL/4I \right) \theta =0$$

Solution

$$\begin{aligned}\text { Given }-& * \text { Restoring tooque } \tau=-\frac{m g L}{2} \sin\theta \\ & * \text { Moment of Inertia }=I\end{aligned} $$
$$\begin{array}{l}\text { Restoring torque- } \\\qquad \tau=-\frac{m g L}{2} \sin \theta \\\text { for very small deflection }\sin\theta\approx\theta \\\Rightarrow \tau=\frac{m g L}{2} \theta \\\text { and }\alpha=\frac{d^{2}\theta}{d t^{2}} \\\Rightarrow\quad I \cdot \frac{d^{2} \theta}{d t^{2}}=\frac{-m g L}{2} \theta \\\Rightarrow\quad\frac{d^{2} \theta}{d t^{2}}+\frac{m g L}{2 I} \theta=0\end{array}$$
Question 9
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Three identical rods, each of length I, are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle is

A
$$\dfrac { \ell }{ \sqrt { 3 } } $$

B
$$\dfrac { \ell }{ \sqrt { 2 } } $$

C
$$\dfrac { \ell }{ \sqrt { 5 } } $$

D
$$\dfrac { \ell }{ \sqrt { 7 } } $$

Solution

$$\begin{array}{l}\text { we know } \\\text { Moment of Inertia of a Rod about its axis } \\\text { and perpendicular is - } \\\qquad I=\frac{M L^{2}}{3} \\\Rightarrow I_{0 A}=I_{0 B}=\frac{M L^{2}}{3} \\\text { * Moment of Inertia of a Rod about its centre } \\\qquad I_{c}=\frac{M L^{2}}{12}\end{array}$$
$$\begin{array}{l}\Rightarrow \text { By parallel axis theorem- } \\\qquad \begin{array}{rl} I_{A B}=I_{c}+M d^{2} & d=\text { distanu from the Axis } \\ I_{A B}=\frac{M L^{2}}{12}+M\left(\frac{\sqrt{3} L}{2} L\right)^{2}=\frac{5 M L^{2}}{6}\end{array} \\\Rightarrow \text { Total moment of Inertia I }=I_{O A}+I_{D B}+I_{A B}=\frac{3 M L^{2}}{2} \\\Rightarrow \text { Radius of gyration (R)- }\end{array}$$
$$ 3 M R^{2}=\frac{3}{2} M L^{2}\quad\Rightarrow\left[R=\frac{L}{\sqrt{2}}\right] \text { option(B) } $$
Question 10
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Force F=300 N acting vertically upward at x=2 m, y=2 m. The magnitude of moment of force about origin is

A
600 Nm

B
660 Nm

C
300 Nm

D
330 Nm

Solution

$$\begin{array}{l}\text { We have force vector }\vec{F}=300 \hat{j} \\\text { and radial vector of point } \vec{\mu}=2 \hat{i}+2 \hat{j} \\\Rightarrow\text { Force moment }=\vec{r}\times\vec{F} \\ =(2 \hat{i}+2 \hat{j}) \times(300 \hat{j}) \\ =600 \hat{k} \\\text { Force moment }\bar{\tau}=600\mathrm{Nm} \\\text { Hence, option }(A) \text { is correct. }\end{array}$$