System of Particles and Rotational Motion Test

Result

System of Particles and Rotational Motion Test - 70

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A ladder $$PQ$$ of length $$5 m$$ inclined to a vertical wall is slipping over a horizontal surface with a velocity of $$2 ms^{-1}$$, when $$Q$$ is at a distance $$3 cm$$ from the ground. Calculate the velocity of centre of mass of the rod at this moment.

A
$$2.667 \ ms^{-1}$$

B
$$1.25 \ ms^{-1}$$

C
$$1.75 \ ms^{-1}$$

D
$$2.75 \ ms^{-1}$$

Solution

In order to find the height of wall when calculated comes to $$3$$.

Through Pythagoras theorem, we come to know that $$x=5$$ and $$y=3$$.

Now with respect to differentiation, the equation comes to $$(2x) \times (dx/dt) + (2 y) \times (dy/dt) = 0$$.

When you substitute $$x$$ and $$y$$, you will know velocity will be $${\dfrac{8}{3}}$$.
Question 2
1 / -0

Two thin rods of mass $$m$$ and length $$l$$, each are joined to form a $$L$$ shape as shown. The moment of inertia of rods about an axis passing through free end (O) of a rod and perpendicular to both the rods is:

A
$$\dfrac{2}{7} ml^2$$

B
$$\dfrac{ml^2}{6} $$

C
$$ml^2$$

D
$$\dfrac{5ml^2}{3} $$

Solution

Moment of inertia of rod 1 about O perpendicular to both the rods
$${ I }_{ 1 }=\dfrac { m{ l }^{ 2 } }{ 12 } +m{ x }^{ 2 }$$ where x is the distance between O and center of mass of rod.
$$l=\sqrt { { l }^{ 2 }+{ (\dfrac { l }{ 2 } ) }^{ 2 } } =\sqrt { \dfrac { { 5l }^{ 2 } }{ 4 } } $$
Moment of inertia of rod 2 about axis through O perpendicular to both rods
$${ I }_{ 2 }=\dfrac { m{ l }^{ 2 } }{ 3 } $$
Total moment of inertia I$$={ I }_{ 1 }{ +I }_{ 2 }$$ -------(1)

$${ I }_{ 1 }=\dfrac { m{ l }^{ 2 } }{ 12 } +\dfrac { 5m{ l }^{ 2 } }{ 4 } =\dfrac { 4m{ l }^{ 2 } }{ 3 } $$
Thus 1 becomes
$$=\dfrac { 4m{ l }^{ 2 } }{ 3 } +\dfrac { m{ l }^{ 2 } }{ 3 } =\dfrac { 5m{ l }^{ 2 } }{ 3 } $$
Question 3
1 / -0

If two particles of masses $$m_1$$ and $$m_2$$ move with velocities $$v_1$$ and $$v_2$$ towards each other on a smooth horizontal plane, what is the velocity of their centre of mass. ?

A
$$V=\dfrac { { m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } $$

B
$$V=\dfrac { { m }_{ 1 }{ v }_{ 1 }-{ m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } $$

C
$$V=\dfrac { { m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }-{ m }_{ 2 } } $$

D
$$V=\dfrac { { m }_{ 1 }{ v }_{ 1 }-{ m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }-{ m }_{ 2 } } $$

Solution

As per center of mass of two objects say 1 and 2 we get then
$$X=\dfrac { { m }_{ 1 }{ x }_{ 1 }+{ m }_{ 2 }{ x }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } $$
$$\text{where}\quad { x }_{ 1 }\quad and\quad { x }_{ 2 }\quad \text{are position of obeject 1 and 2 from reference point respectively }$$
Position of object 1 and 2 at time =t
$${ x }_{ 1 }={ v }_{ 1 }t+{ x }_{ { u }_{ 1 } }\\ { x }_{ 2 }={ v }_{ 2 }t+{ x }_{ { u }_{ 2 } }$$
where $${ x }_{ { u }_{ 1 } }\quad and\quad { x }_{ { u }_{ 2 } }$$ are the initial position of the objects respectively.
Now the center of mass of bjects 1 and 2 at time = t
$$X(t)=\dfrac { \{ { m }_{ 1 }({ v }_{ 1 }t+{ x }_{ { u }_{ 1 } })+{ m }_{ 2 }({ v }_{ 2 }t+{ x }_{ { u }_{ 2 } })\} }{ { m }_{ 1 }+{ m }_{ 2 } } $$
Now differentiating with respect to time t, we get
$$V(t)=\dfrac { d }{ dt } \{ \dfrac { \{ { m }_{ 1 }({ v }_{ 1 }t+{ x }_{ { u }_{ 1 } })+{ m }_{ 2 }({ v }_{ 2 }t+{ x }_{ { u }_{ 2 } })\} }{ { m }_{ 1 }+{ m }_{ 2 } } \} \\ V=\dfrac { { m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } $$
Question 4
1 / -0

A uniform rod AB of length l and mass m is free to rotate about A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A is $$\dfrac{ml^2}{3}$$, the initial angular acceleration of the rod will be?

A
$$\dfrac{3g}{2l}$$

B
$$\dfrac{2g}{3l}$$

C
$$mg\dfrac{l}{2}$$

D
$$\dfrac{3}{2}gl$$

Solution

Given that moment of inertia about A is
$$I={ ml }^{ 3 }/3$$

Now, torque about A is given as
$$\tau=F×r$$
$$\tau=mg\frac { l }{ 2 } $$
$$I\alpha=mg\frac { l }{ 2 } $$
$$\dfrac { { ml }^{ 2 } }{ 3 } \alpha=mg\dfrac { l }{ 2 } $$
$$\alpha=\dfrac { 3g }{ 2l } $$
Question 5
1 / -0

A sphere $$(C.O.M)$$ is moving at $$v_{0}$$ has angular velocity $$\dfrac {3v_{0}}{4\ R}$$ as show. Now it placed on a rough horizontal surface. Its final velocity (of $$C.O.M$$) is:

A
$$\dfrac {v_{0}}{7}$$

B
$$\dfrac {v_{0}}{2}$$

C
$$\dfrac {13v_{0}}{14}$$

D
$$\dfrac {5v_{0}}{7}$$

Solution
Question 6
1 / -0

A body is dropped from a height h, if it acquires a momentum p, then the mass of the body is:

A
$$\dfrac{p}{\sqrt{2gh}}$$

B
$$\dfrac{P^2}{2gh}$$

C
$$\dfrac{2gh}{p}$$

D
$$\sqrt{2gh/p}$$

Solution

momentum (p) = mv
So mass of the body(m) = $$\dfrac{p}{v}$$
$$v = \sqrt{2gh}$$
$$\therefore$$ mass$$ =\dfrac{p}{\sqrt{2gh}}$$
hence (A) is correct answer
Question 7
1 / -0

A small block of mass $$m$$ is moving on a horizontal table surface at initial speed $$v_0$$. It then moves smoothly onto a sloped wedge of mass $$M$$ . The wedge can also move on the table surface. Assume that everything moves without friction. Which of the following statement(s) is (are) correct :-

A
COM of the system always moves rightward

B
COM of the system moves upwards for some time

C
Speed of M always increases during contact with $$ m $$ .

D
Speed of COM of the system is always constant

Solution
Question 8
1 / -0

The centre of mass of non uniform rod of length $$L$$ whose mass per unit length $$\lambda$$ varies as $$\lambda=k{x}^{2}$$ where $$k$$ is a constant and $$x$$ is the distance of any point on rod from left end $$A$$ is (from the same end)

A
at the centre of the rod

B
is at $$x=\cfrac{3L}{4}$$

C
is at $$x=\cfrac{4L}{5}$$

D
is at $$x=\cfrac{5L}{6}$$
Question 9
1 / -0

A thin rod of length $$4l$$, mass $$4$$m is bent at the points as shown in the figure. What is the moment of inertia of the rod about the axis passing through $$O$$ and perpendicular to the plane of the paper?

A
$$\dfrac{ml^2}{3}$$

B
$$\dfrac{10ml^2}{3}$$

C
$$\dfrac{ml^2}{12}$$

D
$$\dfrac{ml^2}{24}$$

Solution

Since, the mass of the rod is $$4 m$$ and the length is $$4 l$$

so, mass of $$AB = BO = OC = CD = m$$

and, length of $$AB = BO = OC = CD = l$$

We know, Moment of inertia of a rod about its end = $${\dfrac{(ml^2)}{3}}$$

Moment of inertia of $$AB$$ about $$B (I_1)$$ = $${\dfrac{ml^2)}{3}}$$

Moment of inertia of $$BO$$ about $$O (I_2)$$ = $${\dfrac{(ml^2)}{3}}$$

Moment of inertia of $$OC$$ about $$O (I_3)$$ = $${\dfrac{(ml^2)}{3}}$$

Moment of inertia of $$CD$$ about $$C (I_4)$$ = $${\dfrac{(ml^2)}{3}}$$

Now, from parallel axis theorem,

Moment of inertia of $$AB$$ about $$O (I_5)$$:

$$ = I_1 + ml^2$$ ————(parallel axis theorem)

$$ = {\dfrac{(ml^2)}{3}} + {ml^2}$$

$$= {\dfrac{(4 ml^2)}{3}}$$

Similarly, Moment of inertia of $$CD$$ about $$O (I_6)$$

$$ = {\dfrac{(4 ml^2)}{3}}$$

So, moment of inertia of the rod about $$O$$

=$$ I_2 + I_3 + I_5 + I_6$$

= $${\dfrac{( ml^2)}{3}} + {\dfrac{( ml^2)}{3}} + {\dfrac{(4 ml^2)}{3}} + {\dfrac{(4 ml^2)}{3}}$$

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

A light rod of length $$l$$ has two masses $$m_1$$ and $$m_2$$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is

A
$$\dfrac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}{l^2}$$

B
$$\dfrac{{{m_1} + {m_2}}}{{{m_1}{m_2}}}{l^2}$$

C
$$\left( {{m_1} + {m_2}} \right){l^2}$$

D
$$\sqrt {{m_1}{m_2}} {l^2}$$

Solution