System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 47

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Question 1
1 / -0

A spherical body of radius $$'R'$$ rolls on a horizontal surface with linear velocity $$'v'$$. Let $$L_1$$ and $$L_2$$ be the magnitudes of angular momenta of the body about centre of mass and point of contact $$P$$. Then,

A
$$L_2 = 2 L_1$$; if radius of gyration $$K = R$$

B
$$L_2 = 2 L_1$$; for all cases

C
$$L_2 > 2 L_1$$; if radius of gyration $$K < R$$

D
None of these

Solution

Angular momentum about center of mass:
$$L_{1}=I\omega = MK^2\omega$$
angular momentum at the point of contact:
$$L_{2}=I\omega + MRv = MK^2\omega + MR(R\omega)$$
$$\because\, (v=R\omega)$$
$$L_{2} = MK^2\omega + MR^2\omega$$
$$L_{2} = M\omega(K^2 + R^2)$$
$$\therefore\, L_{2}= 2L_{1}\, for K=R$$
$$L_{2}>L_{1}\, for K>R$$
Question 2
1 / -0

Two persons A and B of weight 80 kg and 50 kg respectively are standing at opposite ends of a boat of mass 70 kg and length 2m, at rest. When they interchange their positions then the displacement of the center of mass of the boat will be

A
60 cm towards left

B
30 cm towards right

C
30 cm towards left

D
stationary

Solution
Question 3
1 / -0

The mass per unit length of a non-uniform rod of length $$L$$ varies as $$m = \lambda x$$ where $$\lambda$$ is constant. The centre of mass of the rod will be at :

A
$$\dfrac{2}{3}L$$

B
$$\dfrac{3}{2}L$$

C
$$\dfrac{1}{2}L$$

D
$$\dfrac{4}{3}L$$

Solution

$$m=\lambda x$$
taking differentiable on both side
$$dm=\lambda xdx$$
$$x_{cm}=\dfrac{\displaystyle\int^L_0xdm}{\displaystyle\int^L_0dm}$$

$$=\dfrac{\displaystyle\int^L_0xx\lambda dx}{\displaystyle\int^L_0\lambda xdx}$$

$$=\dfrac{\lambda\displaystyle\int^L_0x^2dx}{\lambda\displaystyle\int^L_0xdx}$$

$$=\left[\dfrac{x^3/3}{x^2/2}\right]^L_0$$

$$=\dfrac{L^3}{3}\times \dfrac{2}{L^2}$$

$$x_{cm}=\dfrac{2}{3}L$$.
Question 4
1 / -0

Four identical thin rods each of mass M and length l, form a square frame. Moment of inertia of this frame about an axis through the center of the square and perpendicular to its plane is:

A
$$\cfrac { 1 }{ 3 } M{ l }^{ 2 }$$

B
$$\cfrac { 4 }{ 3 } M{ l }^{ 2 }$$

C
$$\cfrac { 2 }{ 3 } M{ l }^{ 2 }$$

D
$$\cfrac { 15 }{ 3 } M{ l }^{ 2 }$$

Solution

$$Ib{b^1} = Ia{a^1} + m{R^2}$$
from parallel axis theorem
$$b{b^1}$$ is parallel to $$a{a^1}$$
$$Ixd = \dfrac{{M{L^2}}}{{12}}$$
Moment of inertia of frame
$$\begin{array}{l} I=\dfrac { { M{ L^{ 2 } } } }{ { 12 } } =\dfrac { { M{ L^{ 2 } } } }{ 4 } =\dfrac { { M{ L^{ 2 } } } }{ 3 } \\ Total\, \, MOI=\dfrac { { M{ L^{ 2 } } } }{ 3 } +M{ L^{ 3 } }=\dfrac { { 4M{ L^{ 2 } } } }{ 3 } \end{array}$$
Question 5
1 / -0

A body of mass 1 kg has a kinetic energy of motion 8 J. Its linear momentum is equal to:

A
$$ 2 kg ms^{-1} $$

B
$$ 4kg ms^{-1} $$

C
$$ 6kg ms^{-1} $$

D
$$ 8 kg ms^{-1} $$

Solution

We know that,
$$Kl^2=\dfrac{mV^2}{2}$$ and $$P$$ ( linear momentum ) $$=mv$$
So, $$KE=\dfrac{(mV)^2}{2m}=\dfrac{P^2}{2m}$$
$$\Rightarrow P^2=2mKE\Rightarrow P=\sqrt{(2m)(KE)}$$
Here, $$m=1$$ and $$KE=85$$. So,
$$P=\sqrt{2\times (1)(8)}=\sqrt{16}=4\ m\ kg\ s^{-1}$$
Option $$-B$$ is correct.
Question 6
1 / -0

If momenta of two particles of a system are given by $$\overrightarrow { { p }_{ 1 } } =2\hat { i } -\hat { j } +3\hat { k } $$ and $$\overrightarrow { { p }_{ 2 } } =$$ $$-\hat { i } +2\hat { j } +3\hat { k } $$, then the angle made by the direction of motion of the system with $$x$$-axis is

A
$$\cos ^{ -1 }{ 1 } $$

B
$$\cos ^{ -1 }{ \sqrt { 36/38 } } $$

C
$$45^0$$

D
$$\cos ^{ -1 }{ \sqrt { 1/38 } } $$

Solution
Question 7
1 / -0

A $$2$$ kg body and a $$3$$ kg body are moving along the x-axis. At a particular instant the $$2$$ kg body has a velocity of $$3$$ m/s and the $$3$$ kg body has the velocity of $$2$$ m/s. The velocity of the centre of mass at that instant is

A
$$5$$ m/s

B
$$1$$ m/s

C
Zero

D
$$2.4$$ m/s

Solution
Question 8
1 / -0

A T joint is formed by two identical rods A and B each of mass m and length L in the X -Y plane as shown. Its moment of inertia about axis coinciding with A is

A
$$\frac{2mL^2}{3}$$

B
$$\frac{mL^2}{12}$$

C
$$\frac{mL^2}{6}$$

D
None of these

Solution

$$\textbf{Step 1 - Expression for net moment of inertia Refer Figure}$$
$$I_{tot.} = I_{1} + I_{2}$$

$$\textbf{Step 2 - Calculate net moment of inertia}$$
For vertical rod, $$I_{1} = 0$$ because axis passes through the rod.
$$I_{2} = \dfrac {ML^{2}}{12}$$
$$\therefore I_{tot.} = 0 + \dfrac {ML^{2}}{12}$$
$$I_{tot.} = \dfrac {ML^{2}}{12}$$

Hence option B correct.
Question 9
1 / -0

Four thin uniform rods each of length L and mass m are joined to form a square. The moment of inertia of square about an axis along its one diagnol is

A
$$\dfrac{{m{l^2}}}{6}$$

B
$$\dfrac{2}{3}m{l^2}$$

C
$$^{\dfrac{{3m{l^2}}}{4}}$$

D
$$^{\dfrac{{4m{l^2}}}{3}}$$

Solution

Moment of inertia about the centre of one rod, $$I_1=\dfrac{ml^2}{12}$$

Moment of inertia through the centre of the square, will be through an axis parallel to moment of inertia perpendicular to the rod,

$$I_2=I_1+m(\dfrac l2)^2$$

$$I_2=\dfrac{4ml^2}{12}$$

For four rods,

$$I_2=\dfrac{4ml^2}{12}\times 4$$

From perpendicular axis theorem,

$$I_2=I_3+I_4$$

where, $$I_3$$ and $$I_4$$ are moment of inertia through the diagonals. For a square, both diagonals are equal.

$$I_3=I_4$$

$$I_2=2I_3$$

$$\dfrac{4ml^2}{12}\times 4=2I_3$$

$$I_3=\dfrac{2ml^2}{3}$$
Question 10
1 / -0

In free space, a shell moving with velocity 60 m/s along the positive x-axis of an inertial frame, when passes the origin, explodes into two pieces of masses ratio 1 : 2. Velocity of the mass center after the explosion is

A
20 m/s

B
60 m/s

C
90 m/s

D
None of the above

Solution