System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 28

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Question 1
1 / -0

A ladder is leaned against a smooth wall and allowed to slip on a frictionless floor. Which figure represents trace of its COM ?

A

B

C

D
Solution
There are only $$3$$ forces on ladder
$$Mg$$
normal from wall
normal from ground
So the net force will be in $$-\hat{j}$$ due to $$Mg - N$$ and $$-\hat{i}$$due to the normal from the wall.
$$finally$$
the rod will come to rest on the ground so we can see that it has moved down and in left direction so
$$option$$ $$(A)$$ is the correct answer
Question 2
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A body has its center of mass at the origin. The x-axis coordinates of the particles :

A
may be all positive

B
may be all negative

C
should be all at zero

D
may be positive for some case and negative in other cases.

Solution

All co-ordinates positive will make the co-ordinate of com positive.
All co-ordinates negative will make the co-ordinates of com negative.
If all the co-ordinates are zero(0), co-ordinate of com is zero(0).
If co-ordinate of com can be zero, co-ordinates of some are positive and co-ordinates of some are negative.
Question 3
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Two balls of equal masses are projected upward simultaneously, one from the ground with speed $$50m/s$$ and other from a $$40m$$ high tower with initial speed $$30m/s$$. Find the maximum height attained by their centre of mass.

A
100 m

B
200 m

C
150 m

D
50 m

Solution

$${ V }_{ CM }\quad =\quad \dfrac { { m }_{ 1 }{ v }_{ 1 }\quad +\quad { m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }\quad +\quad { m }_{ 2 } } \quad =\quad ({ v }_{ 1 }\quad +\quad { v }_{ 2 })/2\\ $$
$$Initially,\quad { U }_{ CM }\quad =\quad 40\quad m/s\\ { X }_{ CM }\quad =\quad 40/2\quad =\quad 20\quad m\quad above\quad ground.\\ Maximum\quad height\quad means\quad { V }_{ CM }\quad =\quad 0\\ That\quad is,\quad { 0 }^{ 2 }\quad -\quad { 40 }^{ 2 }\quad =\quad 2\times (-10)\times (H-20)\\ Thus\quad H\quad =\quad 100\quad m $$
Question 4
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A particle of mass m is moving in horizontal circle of radius $$r$$ with uniform speed $$v$$. When it moves from one point to a diametrically opposite point its

A
KE changes by $$mv^2$$.

B
KE changes by $$\dfrac{1}{4}mv^2$$.

C
momentum does not change.

D
momentum changes by $$2 mv$$.

Solution

At diametrically opposite points velocity of particle is same but opposite in direction. Hence, change in momentum is
$$ p = mv - (-mv) = mv +mv = 2mv$$
Question 5
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A pulley is hinged at the centre and a massless thread is wrapped around it. The thread is pulled with a constant force F starting from rest. As the time increases:

A
its angular velocity increases, but force on hinge remains constant.

B
its angular velocity remains same, but force on hinge increases.

C
its angular velocity increases and force on hinge increases.

D
its angular velocity remains same and force on hinge is constant.
Solution
- As the pulley is hinged so there is translation equilibrium so net force on pulley is zero.The force on the hinge is constant that is $$F$$ and it is constant.
- Net torque on the pulley is $$R\times F$$ which is constant.
So as the torque on the pulley is constant its angular velocity will keep on increasing with time.
Question 6
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A particle of mass $$m$$ strikes a wall with a velocity $$v$$ making an angle $$\theta$$ with the wall and rebounds.The change in momentum of the particle will be :

A
$$-2m\overline{v} cos \theta$$

B
$$0$$

C
$$2m\overline{v}$$

D
$$-2m\overline{v} sin \theta$$

Solution

the horizontal and vertical components of velocity are $$v sin \theta$$ and $$v cos \theta$$ respectively
after the collision momentum will change only in horizontal direction and not in vertical
initial momentum in horizontal direction is $$m \times v sin \theta$$
after collision velocity will reverse $$m \times -v sin \theta$$
therefore change in momentum is $$\Delta p = p_f-p_i = -mv sin \theta - mv sin\theta = -2 m v sin \theta$$
Question 7
1 / -0

Two identical rods are joined to form the shape of X. the smaller angle between rods is $$\theta$$. The moment of inertia of the system about an axis passing through the point of intersection of the rod and perpendicular to their plane is

A
$$I \propto \theta$$

B
$$I \propto sin^2\theta$$

C
$$I\propto cos^2 \theta$$

D
Independent of $$\theta$$

Solution

The moment of inertia of the two rods will always be the summation of their individual moment of inertias irrespective of the angle between them.
Question 8
1 / -0

The diameter of a solid disc is $$0.5$$ m and its mass is $$16$$ Kg. What torque will increase its angular velocity from zero to $$120$$ rotation/minute in $$8$$ seconds ?

A
$$\displaystyle{\frac{\pi}{4}N/m}$$

B
$$\displaystyle{\frac{\pi}{2}N/m}$$

C
$$\displaystyle{\frac{\pi}{3}N/m}$$

D
$$\pi N/m$$

Solution

Final angular velocity :$$\omega= \dfrac{120 \times 2\pi}{60}=4\pi\: rad/s$$.
Thus its acceleration at the end of the 8th second will be $$\alpha \dfrac{4\pi-0}{8}=\dfrac{\pi}{2}rad/s^2$$. Torque is calculated as
$$\tau =I\alpha=\dfrac{mr^2}{2}\alpha =\dfrac{16((0.25)^2)}{2}\dfrac{\pi}{2}=\pi/4\: Nm$$
Question 9
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A flywheel of moment of inertia $$0.4\,Kg/m^2$$ and radius $$0.2 m$$ is free to rotate about a central axis. If a string is wrapped around it and it is pulled with a force of $$10 N$$ then its angular velocity after $$4 s$$ will be

A
$$5 rad/s$$

B
$$20 rad/s$$

C
$$10 rad/s$$

D
$$0.8 rad/s$$

Solution

The string provides a torque on the flywheel. We have torque as $$\tau= I\alpha$$.
Thus
$$Fr=I\alpha$$
or
$$10\times 0.2=0.4\times \alpha$$
or
$$\alpha=5 rad/s^2$$
Thus after 4 seconds we have the angular velocity as
$$\omega=0+5\times 4$$
$$=20 rad/s$$
Question 10
1 / -0

Two particles of mass 1 kg and 3 kg move towards each other under mutual force of attraction. No other force acts on them. When the relative velocity of approach of the two particles is $$\displaystyle{2 ms^{-1}}$$ their centre of mass has a velocity of $$\displaystyle{0.5 ms^{-1}}$$ and when their velocity of approach become $$\displaystyle{3 ms^{-1}}$$, the velocity of their centre of mass is $$\displaystyle{0.75 ms^{-1}}$$. Then,

A
Above statement is correct

B
Above statement is false

C
Above statement may be correct or incorrect

D
Velocity of centre of mass can change in the given case

Solution

The status of the center of mass cannot be changed by the internal forces. As there is no external force the center of mass will continue to travel with the same velocity of 0.5 m/s, although, the velocity of different particles of the system may change due to the action of internal forces.