System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 37

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Question 1
1 / -0

A small part of the rim of a fly wheel breaks off while it is rotating at a constant angular speed. Then its radius of gyration will

A
Increase

B
Decrease

C
Remain unchanged

D
Nothing definite can be said

Solution

$$I\omega$$ is conserved. If a small piece is broken, $$I$$ will be less and $$\omega$$ will increase But $$I = mk^{2}$$, as $$I$$ is less, $$m$$ is small, $$k^{2}$$ also will be small. The radius of gyration also will decrease.
Question 2
1 / -0

Three identical thin rods, each of mass m and length l are joined to form an equilateral triangle. Find the moment of inertia of the triangle about one of its sides.

A
$$\displaystyle\frac{Ml^2}{2}$$

B
$$\displaystyle\frac{Ml^2}{3}$$

C
$$\displaystyle\frac{Ml^2}{9}$$

D
$$\displaystyle\frac{Ml^2}{12}$$

Solution

$$MOI\quad of\quad system={ I }_{ AB }+{ I }_{ BC }+{ I }_{ AC }\\ { I }_{ AB }=0\quad (it\quad is\quad at\quad the\quad axis)\\ { I }_{ BC }=\dfrac { 1 }{ 3 } m({ l }_{ eff. })^{ 2 }=\dfrac { 1 }{ 3 } m({ l }\sin60)^{ 2 }=\dfrac { 1 }{ 3 } m{ l }^{ 2 }\times \dfrac { 3 }{ 4 } \\ \therefore MOI=0+\dfrac { m{ l }^{ 2 } }{ 4 } +\dfrac { m{ l }^{ 2 } }{ 4 } =\dfrac { m{ l }^{ 2 } }{ 2 } $$
Question 3
1 / -0

A body of mass 1.5 kg rotating about an axis with angular velocity of 0.3 rad $$s^{-1}$$ has the angular momentum of 1.8 kg $$m^2s^{-1}$$. The radius of gyration of the body about an axis is :

A
2 m

B
1.2 m

C
0.2 m

D
1.6 m

E
0.8 m

Solution

We know that,
$$L=l\omega =Mk^2\omega$$ ...(i)$$(\because l=MK^2)$$
Now put the given value in Eq. (i), we get
$$1.8=1.5\times K^2\times 0.3$$
$$K^2=\dfrac{1.8}{1.5\times 0.3}$$
$$K^2=\dfrac{1.8}{0.45}=4$$
$$K=2m$$
Question 4
1 / -0

Two objects $$P$$ and $$Q$$ initially at rest move towards each other under mutual force of attraction. At the instant when the velocity of $$P$$ is $$v$$ and that of $$Q$$ is $$2v$$, the velocity of centre of mass of the system is

A
$$v$$

B
$$3v$$

C
$$2v$$

D
$$1.5v$$

E
Zero

Solution

Since, they move due to mutual interaction between two objects so, centre of mass remains same and its velocity is zero.
Question 5
1 / -0

If a body of moment of inertia $$2$$kg $$m^2$$ revolves about its axis making $$2$$ rotations per school, then its angular momentum(in Js) is:

A
$$2\pi$$

B
$$4\pi$$

C
$$6\pi$$

D
$$8\pi$$

E
$$10\pi$$

Solution

Given,
Moment of inertia of body $$I=2\,kg \,m^2$$
Rotation of body in one second $$f=2$$ rps
We know that,
Angular momentum $$L=l\omega$$
$$L=2\times 2\pi f$$
$$L=2\times 2\pi(2)$$
$$L=8\pi \ J s$$.
Question 6
1 / -0

The radius of gyration of a solid cylinder of mass $$M$$ and radius $$R$$ about its own axis is

A
$$\dfrac { R }{ \sqrt { 2 } } $$

B
$$\dfrac { R }{ 2 } $$

C
$$\dfrac { R }{ \sqrt { 3 } } $$

D
$$\dfrac { R }{ 3 } $$

E
$$\dfrac { R }{ 4 } $$

Solution

The radius of gyration of a solid cylinder
$$K=\sqrt { \dfrac { I }{ M } } =\sqrt { \dfrac { M{ R }^{ 2 } }{ 2 } } $$
$$K=\dfrac { R }{ \sqrt { 2 } } $$
Question 7
1 / -0

A thin rod of length L is suspended from one end and rotated with n rotations per second. The rotational kinetic energy of the rod will be :

A
$$2mL^2\pi^2n^2$$

B
$$\frac{1}{2}mL^2\pi^2n^2$$

C
$$\frac{2}{3}mL^2\pi^2n^2$$

D
$$\frac{1}{6}mL^2\pi^2n^2$$

Solution

Moment of inertia of the rod about the axis passing through its end $$I = \dfrac{mL^2}{3}$$
Frequency of rotation   $$f = n$$
Angular velocity of rod   $$w = 2\pi f = 2\pi n$$
Kinetic energy of the rod   $$K.E = \dfrac{1}{2}Iw^2$$
$$\implies \ K.E. = \dfrac{1}{2}\times \dfrac{mL^2}{3}\times (2\pi n)^2 = \dfrac{2}{3}mL^2 \pi^2 n^2$$
Question 8
1 / -0

The radius of gyration of a solid sphere of radius $$R$$ about a certain axis is $$R$$. The distance of this axis from the centre of the sphere is

A
$$R$$

B
$$\sqrt {0.5R}$$

C
$$\sqrt {0.6R}$$

D
$$\sqrt {0.4R}$$

Solution

Hint:-Use parallel axis theorem to calculate moment of inertia about given axis.

Step 1: Write moment of inertia in terms of radius of gyration and calculate MI about given axis
$$I = mk^{2}$$ (where $$k$$ is the radius of gyration.)
$$I= mR^{2} .... (i)$$
USING PARALLEL AXIS THEOREM
$$I = I_{CM} + mh^{2}$$
$$= \dfrac {2}{5} mR^{2} + mh^{2} ..... (ii)$$

Step :2 Solving equations
From (i) and (ii)
$$mR^{2} = \dfrac {2}{5}mR^{2} + mh^{2}$$
$$\dfrac {3}{5}mR^{2} = mh^{2} \Rightarrow h^{2} = \dfrac {3}{5} R^{2}$$
$$\therefore h = \sqrt {0.6R}$$.
Question 9
1 / -0

A rod of mass $$m$$ and length $$L$$, lying horizontally, is free to rotate about a vertical axis through its centre. A horizontal force of constant magnitude $$F$$ acts on the rod at a distance of $$L/4$$ from the centre. The force is always perpendicular to the rod. Find the angle rotated by the rod during the time $$t$$ after the motion starts.

A
$$\dfrac{3Ft^2}{5mL}$$

B
$$\dfrac{5Ft^2}{2mL}$$

C
$$\dfrac{3Ft^2}{2mL}$$

D
$$\dfrac{2Ft^2}{3mL}$$

Solution

We have

torque $$=\tau=rF\sin { 90° } =\cfrac{ LF }{ 4 } $$

And perpendicular axis theorem gives us

$$I={ I }_{ 2 }={ I }_{ x }+{ I }_{ y }=\cfrac{ M{ L }^{ 2 } }{ 12 } $$

Angular acceleration,

$$\alpha =\cfrac{ \tau }{ I } =\cfrac{ LF }{ 4 } \times \left( \cfrac{ 12 }{ M{ L }^{ 2 } } \right) =\cfrac{ 3F }{ ML } $$

If $$\theta$$ is the angle rotated in time $$t$$, and initial angular velocity $${ w }_{ 0 }$$being zero we have

$$\theta ={ w }_{ 0 }t+\cfrac{ 1 }{ 2 } \alpha { t }^{ 2 }=\cfrac{ 3F{ t }^{ 2 } }{ 2ML } $$
Question 10
1 / -0

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

A
$$zero$$

B
$$v$$

C
$$1.5v$$

D
$$3v$$

Solution

Zero, according to newton law of motion if body is at rest and no external force is acting then body will be at rest until external force acts.
As we know both of these are moving due to their mutual attraction so velocity of center of mass will be constant ,i.e. $$zero$$.

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Re-Attempt Weekly Quiz Competition

System of Particles and Rotational Motion Test - 37

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System of Particles and Rotational Motion Test - 37

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Question 1

Increase

Decrease

Remain unchanged

Nothing definite can be said

Solution

Question 2

$$\displaystyle\frac{Ml^2}{2}$$

$$\displaystyle\frac{Ml^2}{3}$$

$$\displaystyle\frac{Ml^2}{9}$$

$$\displaystyle\frac{Ml^2}{12}$$

Solution

Question 3

2 m

1.2 m

0.2 m

1.6 m

0.8 m

Solution

Question 4

$$v$$

$$3v$$

$$2v$$

$$1.5v$$

Zero

Solution

Question 5

$$2\pi$$

$$4\pi$$

$$6\pi$$

$$8\pi$$

$$10\pi$$

Solution

Question 6

$$\dfrac { R }{ \sqrt { 2 } } $$

$$\dfrac { R }{ 2 } $$

$$\dfrac { R }{ \sqrt { 3 } } $$

$$\dfrac { R }{ 3 } $$

$$\dfrac { R }{ 4 } $$

Solution

Question 7

$$2mL^2\pi^2n^2$$

$$\frac{1}{2}mL^2\pi^2n^2$$

$$\frac{2}{3}mL^2\pi^2n^2$$

$$\frac{1}{6}mL^2\pi^2n^2$$

Solution

Question 8

$$R$$

$$\sqrt {0.5R}$$

$$\sqrt {0.6R}$$

$$\sqrt {0.4R}$$

Solution

Question 9

$$\dfrac{3Ft^2}{5mL}$$

$$\dfrac{5Ft^2}{2mL}$$

$$\dfrac{3Ft^2}{2mL}$$

$$\dfrac{2Ft^2}{3mL}$$

Solution

Question 10

$$zero$$

$$v$$

$$1.5v$$

$$3v$$

Solution

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