System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 80

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Weekly Quiz Competition

Question 1
1 / -0

A disk and a ring of the same mass are rolling to have the same kinetic energy. What is ratio of the velocities of centre of mass

A
$$( 4 : 3 ) ^ { 1 / 2 }$$

B
$$( 3 : 4 ) ^ { 1 / 2 }$$

C
$$( 2 ) ^ { 1 / 2 } : ( 3 ) ^ { 1 / 2 }$$

D
$$( 3 ) ^ { 1 / 2 } : ( 2 ) ^ { 1 / 2 }$$

Solution
Question 2
1 / -0

A 1 m long rod has mass of $$ o.12 kg $$ what is the moment of inertia about an axis passing through the centre and perpendicular to the length of rod

A
$$0.01kg-{ m }^{ 2 }$$

B
$$0.001 kg-{ m }^{ 2 }$$

C
$$1 kg-{ m }^{ 2 }$$

D
$$10 kg-{ m }^{ 2 } $$

Solution
Question 3
1 / -0

If a disc of mass $$m$$ and radius $$r$$ is reshaped into a ring of radius $$2$$ $$r$$ , the mass remaining the same, the radius of gyration about centroidal axis perpendicular to plane goes up by a factor of

A
$$\sqrt { 2 }$$

B
$$2$$

C
$$2$$$$\sqrt { 2 }$$

D
$$4$$

Solution
Question 4
1 / -0

An electric fan has blades of length $$30 cm$$ measured from the axis of rotation. If the fan is running at $$120 rpm$$, the acceleration of a point on the tip of the blade is

A
$$1600 ms^{ -2 }$$

B
$$47.4 ms^{ -2 }$$

C
$$23.7 ms^{ -2 }$$

D
$$50.5 ms^{ -2 }$$

Solution
Question 5
1 / -0

Mass of thin long metal rod is 2 kg and its moment of inertia about an axis perpendicular to the length of rod and passing through its one end is $$0.5 kgm^2$$. Its radius of gyration is

A
20 cm

B
40 cm

C
50 cm

D
1 m

Solution
Question 6
1 / -0

A body of mass 'm' falls from a height "h' and rebounds. If'e' is the coefficient of restitution between the ground and the body, the change in linear momentum of the body is

A
$$( 2 m e ) \sqrt { 2 g h }$$

B
$$( \mathrm { me } ) \sqrt { 2 \mathrm { gh } }$$

C
$$\mathrm { m } ( 1 + \mathrm { e } ) \sqrt { 2 \mathrm { gh } }$$

D
$$m ( 1 - e ) \sqrt { 2 g h }$$

Solution
Question 7
1 / -0

A semicircular portion of radius $$r$$ is cut from a uniform rectangular plates as shown in figure. The distance of centre of mass $$C$$ of remaining plate, from point $$O$$ is:

A
$$\dfrac{2r}{\left(3-\pi\right)}$$

B
$$\dfrac{3r}{2\left(4-\pi\right)}$$

C
$$\dfrac{2r}{3\left(4+\pi\right)}$$

D
$$\dfrac{2r}{3\left(4-\pi\right)}$$

Solution
Question 8
1 / -0

Two particles $$ A $$ and $$ B $$ initially at rest, move towards each other under the mutual force of attraction. At the instant when the speed of $$ A $$ is $$ v $$ and the speed of $$ B $$ is $$ 2 v, $$ the speed of the centre of mass of the system is:-

A
$$3v$$

B
$$v$$

C
$$1.5v$$

D
$$zero$$

Solution

Given,
velocity of one particle $$A$$ is $$v$$
velocity of another particle $$B$$ is $$2v$$
We know that
$$Center\quad of\quad mass=\frac { { m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } $$
Here there is no external force therefore,
$${ F }_{ ext }=0$$
No change in momentum in center of mass hence, center of mass cannot be zero but velocity may be zero hence, velocity of center of mass will be zero.
Question 9
1 / -0

An arrow sign is made by cutting and rejoining a quarter part of a square plate of side 'L' shown. The distance OC, where 'C' is the centre of mass of the arrow, is

A
$$\dfrac { L }{ 3 } $$

B
$$\dfrac { L }{ 4 } $$

C
$$\dfrac { 3L }{ 8 } $$

D
None of these

Solution

Correct Answer: Option B

Step 1: Mass of the part removed
Let the mass of the square be $$M$$
The removed triangle is one fourth of the total square
$$\therefore M_t = \dfrac M4 $$

Step 2: Center of mass of different parts
The center of mass of the square lies at $$x_1$$ that is $$O$$
Let $$O$$ be the origin.
Refer figure $$1$$
Let the initial position of the triangle be termed as $$2$$ and a triangle has its center of mass at $$\dfrac{2L}{3}$$ from the apex.
so, $$x_{2} = \dfrac{-2}{3} \times \dfrac L2 = - \dfrac L3$$

Let the final position of the triangle be termed as $$3$$.
Now, the center of mass of the triangle in position $$2$$ is $$x_{3} = \dfrac L2 + \dfrac L2 \times \dfrac 13 = \dfrac{2L}{3}$$

Step 3: Center of mass of the system
We will use the negative mass concept
As the triangle is removed from position $$1$$, let its mass be $$M_2 = -\dfrac M4$$
And then it is added at position $$2$$ so mass $$M_3$$ is $$\dfrac M4$$
we know
$$x_{com} = \dfrac{M_1x_1 + M_2x_2 + M_3x_3}{M_1 + M_2 + M_3}$$
so,
$$x_{com} = \dfrac{M\times 0 + (- \dfrac M4) \times -\dfrac{L}{3} + \dfrac M4 \times \dfrac{2L}{3}}{M + (-\dfrac M4) + \dfrac M4}$$

$$\therefore x_{com} = \dfrac L4$$

Hence, correct answer is option B.
Question 10
1 / -0

If the linear density of a rod of length L varies as $$\lambda =A+B_x$$, compute its centre of mass.

A
$$[\cfrac {L(3A+2BL)}{3(2A+BL},0,0]$$

B
$$[0,\cfrac {(3A+2B)L}{(2A+3L},\cfrac L 2]$$

C
$$[0,0\cfrac {L(3A+2BL)}{3(2A+BL}]$$

D
$$[\cfrac L 2,00]$$

Solution