System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 17

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Question 1
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The coordinates of centre of mass of a uniform flag shaped lamina (thin flat plate) of mass $$4kg$$. (The coordinates of the same are shown in figure) are:

A
$$(0.75m, \, 0.75m)$$

B
$$(1.25m, \, 1.50m)$$

C
$$(1m, \, 1.75m)$$

D
$$(0.75m, \, 1.75m)$$

Solution

The lamina can be slit into two rectangular parts.

Mass of part $$A = \sigma \times A_A$$

$$= \sigma \times 2 \times 1 = 2\sigma$$

Mass of part B = \sigma \times A_B$$

$$=\sigma \times 1 \times 2 = 2\sigma$$

Coordinates of the center of masses of the rectangular parts are given by $$x_{Acom} = \dfrac{x_1 + x_2}{2}$$ and $$y_{Ac.o.m} = \dfrac{y_1 + y_2}{2}$$

For part $$A$$,

$$x_{c.o.m} = \dfrac{0+2}{2} = 1$$, $$y_{c.o.m} = \dfrac{2+3}{2} = \dfrac{5}{2}$$

For part $$B$$,

$$x_{c.o.m} = \dfrac{0+1}{2} = \dfrac{1}{2}$$, $$y_{c.o.m} = \dfrac{2+0}{2} = 1$$

Now combining both parts, the center of mass will have the coordinates

$$x_{c.o.m} = \dfrac{m_Ax_A + m_Bx_B}{m_A+m_B} = \dfrac{2\sigma(1) + 2\sigma(\frac{1}{2}) }{2\sigma + 2 \sigma}$$

$$= \dfrac{3\sigma}{4\sigma} = \dfrac{3}{4}$$

$$y_{c.o.m} = \dfrac{m_Ay_A + m_By_B}{m_A + m_B} = \dfrac{2\sigma(\frac{5}{2} )+ 2\sigma(1)}{2\sigma + 2\sigma}$$

$$= \dfrac{7\sigma}{4\sigma} = \dfrac{7}{4}$$

$$\therefore $$ the required co-ordinate is $$\left(\dfrac{3}{4}, \dfrac{7}{4}\right)$$ or $$(0.75, 1.75)$$
Question 2
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A body of mass $$\mathrm{m}=3.513$$ kg is moving along the $$\mathrm{x}$$-axis with a speed of 5.00 $$\mathrm{m}\mathrm{s}^{-1}$$. The magnitude of its momentum is recorded as :

A
17.56 kg $$\mathrm{m}\mathrm{s}^{-1}$$

B
17.57 kg $$\mathrm{m}\mathrm{s}^{-1}$$

C
17. 6 kg $$\mathrm{m}\mathrm{s}^{-1}$$

D
17.565 kg $$\mathrm{m}\mathrm{s}^{-1}$$

Solution

Momentum of the body is $$mv=3.513kg\times 5.00m/s=17.565kg m/s$$
However, the accuracy of the result would be determined by the most inaccurate observation, which is speed with three significant digits. Thus the answer would be expressed in three significant digits, that is, $$17.6\ kg m/s$$
Question 3
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A rigid massless rod of length 3$$l$$ has two masses attached art each end as shown in the figure. The rod is pivoted at point P on the horizontal position, its instantaneous angular acceleration will be:

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

B
$$\dfrac{7g}{3l}$$

C
$$\dfrac{g}{13l}$$

D
$$\dfrac{g}{3l}$$

Solution
Question 4
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Directions For Questions

Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is $$x(t)$$ vs. $$p(t)$$ curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative.

...view full instructions

The phase space diagram for a ball thrown vertically up from ground is

A

B

C

D

Solution

At a particular time say $$t = 0$$s
Particle has velocity $$+v$$
and thus momentum $$p(t= 0) = +mv$$
Also, displacement is 0
At time $$t$$,
Particle reaches maximum height with displacement $$x = h$$
Here, velocity $$v = 0$$
Now,
when particle comes back to its initial position,
Velocity is reversed i.e. $$v = -v$$
and thus,
momentum, $$p = -mv$$
So, we get graph corresponding Option D.
Question 5
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(i) Centre of gravity (C.G.) of a body is the point at which the weight of the body acts
(ii) Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius
(iii) To evaluate the gravitational field intensity due to any body at an external point, the entire mass of the body can be considered to be concentrated at its C.G.
(iv) The radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the C.G. of the body to the axis
Which one of the following pairs of statements is correct ?

A
(iv) and (i)

B
(i) and (ii)

C
(ii) and (iii)

D
(iii) and (iv)

Solution

Centre of gravity (C.G.) of a body is the point at which the weight of the body acts
The radius of gyration of any body rotating about an axis is the length of the perpendicular dropped from the C.G. of the body to the axis
Question 6
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Which of the following statements are correct?
(a) Centre of mass of a body always coincides with the centre of gravity of the body.
(b) Centre of mass of a body is the point at which the total gravitational torque on the body is zero.
(c) A couple on a body produce both translational and rotational motion in a body.
(d) Mechanical advantage greater than one means that small effort can be used to lift a large load.

A
(c) and (d)

B
(b) and (d)

C
(a) and (b)

D
(b) and (c)

Solution

Centre of mass:
It is the point where whole mass of the body is concentrated. If an external force is applied on COM , it torque will be zero.
Centre of gravity:
It is the point where the torque of gravitational force is zero.
COM and CAG mag or mag not coincides
couple only produces rotation, no translation
Mechanical advantage $$=\dfrac{\text{output force}}{\text{Input force}}$$
If it is grater than 1
Output > 1 input
So, less effort is required to lift large low
Question 7
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A stone is dropped from a height $$h$$. It hits the ground with a certain momentum $$P$$. If the same stone is dropped from a height $$100$$% more than the previous height, the momentum when it hits the ground will change by (approximately) :

A
$$41$$%

B
$$53$$%

C
$$60$$%

D
$$68$$%

Solution

Velocity, $$v=\sqrt {2gh}$$

Momentum, $$P=mv=m\sqrt {2gh}$$

$$P\propto \sqrt {h}$$

$$\cfrac{{P}_{2}}{{P}_{1}}=\sqrt {\cfrac{{h}_{2}}{{h}_{1}}}=\sqrt {\cfrac{2h}{h}}=\sqrt 2$$

$${P}_{2}=1.414{P}_{1}$$

% change $$=\cfrac{{P}_{2}-{P}_{1}}{{P}_{1}}\times 100$$%

$$\dfrac{1.414P_1-P_1}{P_1}\times 100$$

% change $$=41.4$$%
Question 8
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The force 'F' acting on a particle of mass 'm' is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8 s is

A
24 Ns

B
20 Ns

C
12 Ns

D
6 Ns

Solution

$$\Delta P=\left (\dfrac {1}{2}\right )(2)(6)-(3)(2)+(4)(3)$$
$$=6-6+12$$
$$=12$$
Question 9
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Two spheres of same size one of mass $$2 kg$$ and another of mass $$4 kg$$ are dropped simultaneously from the top of Qutab Minar (height $$= 72 m$$). When they are $$1 m$$ above the ground, the two spheres have the same:

A
momentum

B
kinetic energy

C
potential energy

D
acceleration

Solution

Momentum , Kinetic energy and Potential energy has the term of mass in them and since they have different mass but same velocity all these quantity would be different but since acceleration is always 'g' it will be same for both.
Question 10
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A thin rod of length $$4l$$ and $$4m$$ is bent at the points as shown in figure. What is the moment of inertia of the rod about the axis passes through point $$O$$ and perpendicular to the plane of paper?

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

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

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

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

Solution

Total moment of inertia
$$= I_{1} + I_{2} + I_{3} + I_{4} = 2I_{1} + 2I_{2}$$
$$= 2(I_{1} + I_{2}) [I_{3} + I_{1}, I_{1} = I_{4}]$$
Now, $$I_{2} = I_{3} = \dfrac {MI^{2}}{3}$$
Using parallel axes theorem, we have
$$I = I_{CM} + Mx^{2}$$ and $$x = \sqrt {l^{2} + \dfrac {l^{2}}{4}}$$
$$I_{1} = I_{4} = \dfrac {Ml^{2}}{12} + M \left [\sqrt {l^{2} + \left (\dfrac {1}{2}\right )^{2}}\right ]^{2}$$
Putting all values we get
Moment of inertia, $$I = 10\left (\dfrac {MI^{2}}{3}\right )$$.