System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 68

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Question 1
1 / -0

Two uniform thin rods each of mass $$'m'$$ and length $$'L'$$ are arranged to form a cross. The moment of inertia of the system about an angular bisector is

A
$$\dfrac { m{ L }^{ 2 } }{ 12 } $$

B
$$\dfrac { m{ L }^{ 2 } }{ 6 } $$

C
$$\dfrac { m{ L }^{ 2 } }{ 3 } $$

D
$$\dfrac { 2m{ L }^{ 2 } }{ 3 } $$

Solution

Moment of inertia of the rod about its centre is I=$$\dfrac{ML^2}{12}$$
Moment of inertia of the system of rods which form a cross about an axis passing through the centre and perpendicular to the rods is I=$$I_x+I_y=2I$$=$$\dfrac{2ML^2}{12}$$
Moment of inertia of the system through angular bisectors of rods = $$\dfrac{I}{2}=\dfrac{ML^2}{12}$$
Question 2
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The radius of gyration of a body about an axis passing through its centre of mass is $$24\ cm$$. Calculate the radius of gyration of the body about parallel axis passing through a point at a distance $$7\ cm$$ from its centre of mass.

A
$$16\ cm$$

B
$$20\ cm$$

C
$$30\ cm$$

D
$$25\ cm$$

Solution
Question 3
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The radius of gyration of a body about an axis at a distance of $$4\ cm$$ from its centre of mass is $$5\ cm$$. The radius of gyration about a parallel axis through centre of mass is:

A
$$2\ cm$$

B
$$5\ cm$$

C
$$4\ cm$$

D
$$3\ cm$$

Solution

$$mr_{gCM}^2+mr^2=mr_{g}^2$$
$$r_{gCM}^2+r^2=r_{g}^2$$
$$r_{gCM}^2+4^2=5^2$$
$$r_{gCM}=3cms$$
Question 4
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The ratio of the radii of gyration of a spherical shell and a solid sphere of the same mass and radius about a tangential axis is:

A
$$\sqrt { 3 } :\sqrt { 7 }$$

B
$$\surd 5:\surd 6$$

C
$$\sqrt { 25 } :\sqrt { 21 }$$

D
$$\sqrt { 21 } :\sqrt { 25 }$$

Solution

Moment of inertia of a solid sphere about tangential axis is $$\dfrac{2}{3}mr^2+mr^2=\dfrac{5}{3}mr^2$$
Radius of gyration=$$\sqrt{\dfrac{5}{3}}r$$
Moment of inertia of a spherical shell about tangential axis is $$\dfrac{2}{5}mr^2+mr^2=\dfrac{7}{5}mr^2$$
Radius of gyration=$$\sqrt{\dfrac{7}{5}}r$$
Ratio of their radii of gyration=$$\sqrt{\dfrac{3}{7}}$$
Question 5
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A thin rod of length $$0.6m$$ is vertically straight on the horizontal floor. This falls freely to one side without slipping of its bottom. The angular velocity of the centre of a rod when its top end touches the floor is?

A
$$5\sqrt { 2} rad$$

B
$$\sqrt { \dfrac { 3}{ 2 } }rad$$

C
$$\sqrt { 3 } rad$$

D
$$\sqrt { \dfrac { 3 }{ 4 } }rad$$

Solution
Question 6
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Two identical discs of mass $$M$$ and radius $$R$$ are joined to form a figure of eight [see Figure]. The radius of gyration about an axis through their point of contact and perpendicular to the plane:

A
$$\sqrt { 5 }\ R$$

B
$$\sqrt { 3 }\ R$$

C
$$\sqrt { 7 }\ R$$

D
$$\sqrt { \dfrac { 3 }{ 2 } }\ R$$

Solution
Question 7
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A uniform rod is kept vertically on a horizontal smooth surface at a point $$O$$. If it is disturbed slightly and released, it falls down on the horizontal surface. The lower end will be

A
At $$O$$

B
At a distance less than $$l/2$$ from $$O$$

C
At a distance $$l/2$$ from $$O$$

D
At a distance larger than $$l/2$$ from $$O$$

Solution
Question 8
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Momentum of inertia of a rod of mass $$2\ kg$$ and length $$1\ m$$ about an axis passing through a point $$25\ cm$$ from the center and normal to the length is:

A
$$2.9\ kg\ {m}^{2}$$

B
$$1.9\ kg\ {m}^{2}$$

C
$$0.29\ kg\ {m}^{2}$$

D
$$0.19\ kg\ {m}^{2}$$

Solution

Moment of inertia=$$\dfrac{ml^2}{12}+mr^2$$
$$\dfrac{2\times 1^2}{12}+2\times (\dfrac{1}{4})^2$$=0.29kgm²
Question 9
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A rod is placed along the line $$y=2\ x$$ with its centre at origin. The moment of inertia of the rod is maximum about

A
$$x-$$axis

B
$$y-$$axis

C
$$z-$$axis

D
Date insufficient

Solution
Question 10
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A uniform disc of mass m and radius R is projected horizontally with velocity $$v_0$$ on a rough horizontal floor, so that it starts. off with a purely sliding motion at t = 0. After $$t_0$$ seconds, it acquires a purely rolling motion as shown in figure. The velocity of the centre of mass of the disc at $$t_0$$.

A
$$\frac{3}{2}v_0$$

B
$$\frac{3}{4}v_0$$

C
$$\frac{2}{3}v_0$$

D
$$\frac{4}{3}v_0$$

Solution