Classical Mechanics MCQ Quiz in मल्याळम - Objective Question with Answer for Classical Mechanics - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

• Given the transformations \(X=\frac{α p}{x}\) and \(P = βx²\), we have to check them against the requirements of canonical transformations.
• The main requirement, among others, is that the Poisson bracket {X, P} = 1.
• In terms of partial derivatives, the Poisson bracket can be written as: \({X, P} = (\frac{∂X}{∂x})(\frac{∂P}{∂p}) - (\frac{∂X}{∂p})(\frac{∂P}{∂x})\)
• Substituting X and P from the problem statement, we get: \({X, P} = (\frac{∂(αp/x)}{∂x})(\frac{∂(βx²)}{∂p}) - (\frac{∂(αp/x)}{∂p})(\frac{∂(βx²)}{∂x})\)
• Computing partial derivatives, we get:\({X, P} = (\frac{-αp}{x²})(0) - (\frac{α}{x})(2βx) = -2αβ \)
• Setting the above equation to 1: \(-2αβ = 1\)
• Finally, \(1+2αβ = 0\)

Explanation:

In this case, we use the formula for the precession of an axially symmetric freely rotating rigid body, caused by the torque resulting from external forces (such as gravity) trying to alter the tilt of the rotation axis. This forms what is known as a gyroscope.

The formula for the precession rate (angular frequency) of a freely spinning body is given by:

 \(ω_{precession} = ω_0 × {(\frac{ I_{axis} }{I_{perpendicular}}-1)}\)

where

• \(ω_0\) is the rotational angular frequency of the earth about its axis of symmetry,
• \(I_{perpendicular}\) is the moment of inertia of the earth about an axis perpendicular to the axis of symmetry,
• \(I_{axis}\) is the moment of inertia of the earth about the axis of symmetry.

Solving for the time period of precession, which is the reciprocal of the precession rate, we get: \(T_{precession} =T_0\times\frac1{(\frac{ I_{axis} }{I_{perpendicular}}-1)}\)

It is given that the ratio of the moment of inertia about the axis of symmetry \(I_{axis}\) to the moment of inertia about an axis perpendicular to this \(I_{perpendicular}\) is 33:32, which makes: \(\frac {I_{axis}}{ I_{perpendicular}} = \frac {33}{32}\)

Substituting this into the precession period formula, we get: \(T_{precession} = T_0 × \frac {1 }{( \frac {33}{32} - 1)}\) 

\(T_{precession} = 32 T_0\)

Explanation:

We can use Newton's second law of motion to solve this problem.

The equation of motion for the particle is given by:

\(m \dfrac{d^2x}{dt^2} = -kx^3 \ \)

We can rewrite this equation as:

\(\dfrac{d^2x}{dt^2} = -\dfrac{k}{m}x^3 \ \ \)

We can multiply both sides of this equation by \(\dfrac{dx}{dt}\ \) to obtain:

\(\dfrac{d}{dt} \left(\dfrac{1}{2} \left(\dfrac{dx}{dt}\right)^2\right) = -\dfrac{k}{m}x^3 \dfrac{dx}{dt} \ \).

We can integrate both sides of this equation with respect to time to obtain:

\(\dfrac{1}{2} \left(\dfrac{dx}{dt}\right)^2 = \dfrac{k}{4m} (a^4 - x^4) \ \)

At the equilibrium position x = 0, we have:

\(\dfrac{1}{2} \left(\dfrac{dx}{dt}\right)^2 = \dfrac{k}{4m} a^4 \ \)

Therefore, the velocity of the particle when it passes through its equilibrium position is:

\(\left|\dfrac{dx}{dt}\right| = \sqrt{\dfrac{k}{2m}a^2} . \ \)

Therefore, the correct option is (2).

Concept:

Phase - space diagram or phase plot is a plot between generalized coordinate x and generalized momentum p.

Calculation:

The Lagrangian and Hamiltonian for the system is given as follows:

L = T - V

= \({1\over 2} m \dot{x}^2 - mgx \)

H = \({p^2 \over 2m} + mgx \) 

E = \(\rm\frac{p^{2}}{2m}\) + mgx

⇒ p² = 2m(E − mgx) which is equation of parabola.

At x = 0,

p = \(\pm \sqrt{2mE} \)

At p = 0,

x = \({E \over mg}\)

For x < 0, when the potential energy is ∞ then the kinetic energy will be 0.

The correct answer is option (2).

Concept:

Lagrangian mechanics enables us to find the equations of motion when the Newtonian method is proving difficult.

In Lagrangian mechanics we start, as usual, by drawing a large, clear diagram of the system, using a ruler and a compass.

Explanation:

As φ is cyclic coordinate, so \(\rm\frac{\partial L}{\partial \dot{\varphi}}\) = pφ =\(m l^2 \sin^2 \theta \dot{\phi} \), is a constant since m, l and g are constants.

Now using the Euler Lagrangian equation for motion we get:

\(\frac{d}{dt} (\frac{\partial L}{\partial \dot{\phi}}) =0 \).

Thus,  \(m l^2 \sin^2 \theta \dot{\phi} \) is conserved.

The correct answer is option (1).

Explanation:

We known that:

P = \(\rm\frac{m v}{\sqrt{1−\frac{v^{2}}{c^{2}}}}\) ⇒ F = \(\rm\frac{d P}{d t}\) = \(\rm m\frac{d v}{d t} \cdot \frac{1}{\sqrt{1−\frac{v^{2}}{c^{2}}}}+m v\left(−\frac{1}{2}\right) \cdot \frac{1}{\left(1−\frac{v^{2}}{c^{2}}\right)^{3/2}} \cdot \frac{−2 v}{c^{2}} \frac{d v}{d t}\)

⇒ F = \(\rm m \frac{d v}{d t} \frac{1}{\sqrt{1−\frac{v^{2}}{c^{2}}}}\left(1+\frac{1}{2} \frac{v^{2}/c^{2}}{\left(1−\frac{v^{2}}{c^{2}}\right)}\right) = m \frac{d v}{d t}\left(\frac{1−v^{2}/2 c^{2}}{\left(1−\frac{v^{2}}{c^{2}}\right)^{3/2}}\right)\)

= \(\rm m\frac{d v}{d t}\left[\frac{\left(1−v^{2}/c^{2}\right)^{1/2}}{\left(1−v^{2}/c^{2}\right)\left(1−v^{2}/c^{2}\right)^{1/2}}\right] = \frac{m c^{2}}{\left(c^{2}−v^{2}\right)} \frac{d v}{dt}\)

The correct option is option (3).

Concept:

Lagrange's equation of motion is a mathematical equation that is used to describe the motion of a physical system in terms of its Lagrangian function.

The Lagrangian function is a mathematical function that describes the energy of a system in terms of its position and velocity.

Explanation:

The Euler equation of motion is given as:

\(\rm\frac{d}{dt}\left(\frac{d L}{d \dot{x}}\right)=\frac{dL}{dx}\) ⇒ mẍ(1 + 4a2x2) = −4ma2xẋ2 − 2mgax.

The correct option is option (2).

Explanation:

Equation of constrain is given by y = ax2, K.E., T = \(\frac{1}{2}\)m(ẋ2 + ẏ2)

ẏ = 2axẋ ⇒ T = \(\frac{1}{2}\)m(ẋ2 + 4a2x2ẋ2) = \(\frac{1}{2}\)mẋ2(1 + 4ax2)

V = mgy = mgax2.

∵ L = T − V ⇒ L = \(\frac{1}{2}\)m(1 + 4a2x2)ẋ2 − mgax2   

The correct option is option (2).

Concept:

The Lagranges equation of motion of a system is given by

\({d \over dt} {\partial L \over \partial \dot{q}} - {\partial L \over \partial q} = 0\)

Calculation:

The Lagrangian L depends on 

L(x,y,\(̇ x\),\(̇ y\))

\({d \over dt} {\partial L \over \partial \dot{x}} - {\partial L \over \partial x} = 0\)

\({d \over dt} {\partial L \over \partial \dot{ y}} - {\partial L \over \partial y} = 0 \)

L^' = L(x,y,\(\dot x\),\(\dot{y}\))

\({d' \over dt'} {\partial L' \over \partial x} - {\partial L' \over \partial x} = ​​{d \over dt} {\partial L \over \partial x} - {\partial L \over \partial x}+ \ddot{y} = 0\)

⇒\(\dot{y} = c_1\)

Similarly \(\dot{x} = c_2\)

The correct answer is option (2).

CONCEPT:

As we know,

\(\vec F = - \vec \nabla V\)

CALCULATION:

Given: \({\rm{V}}\,{\rm{ = }}\,\frac{{\rm{1}}}{{\rm{2}}}{\rm{m}}\left( {{\rm{\omega }}_{\rm{1}}^{\rm{2}}{{\rm{x}}^{\rm{2}}}\,{\rm{ + }}\,{\rm{\omega }}_{\rm{2}}^{\rm{2}}{{\rm{y}}^{\rm{2}}}\,{\rm{ + }}\,{\rm{\omega }}_{\rm{3}}^{\rm{2}}{{\rm{z}}^{\rm{2}}}} \right)\)

As we have;

\(\vec F = - \vec \nabla V\)

⇒ \(\vec F = - m(\omega_1^2 x \hat i +\omega_2^2 y \hat j +\omega_3^2 z \hat k) \)

Now the centrifugal force on the particle is written as;

\(\vec F_{cf} =-m\vec\omega\times (\vec \omega \times \vec r)\)

Here, \(\vec r = x \hat i+y\hat j+z\hat k\)  and \(\vec \omega = \hat k \Omega \)

so we have;

\(\vec \omega \times \vec r = \Omega (y \hat i-x \hat j )\)

⇒ \(\vec F_{cf} = -m\vec\omega\times \Omega (y \hat i-x \hat j )\)

⇒ \(\vec F_{cf} = -m\Omega^2[\hat k\times (y \hat i-x \hat j )]\)

⇒\(\vec F_{cf} = m\Omega^2(x \hat i+y \hat j)\)

Since, \(\vec F > \vec F_{cf}\)

\(- m(\omega_1^2 x \hat i +\omega_2^2y \hat j +\omega_3^2 z \hat k) >-m\Omega^2(x \hat j+y \hat i)\)

Now, compare the x,y, and z components we have;

\(\omega_1^2x >\Omega ^2 x\)

\(\Omega^2 - \omega_1^2 <0\)

Similarly, \(\Omega^2 - \omega_2^2 <0\)

and \(\omega_3^2 >0\)

Combining we have;

\(\left( {{\Omega ^2}\, - \,{\rm{\omega }}_1^2} \right)\,\left( {{\Omega ^2}\, - \,{\rm{\omega }}_2^2} \right) < 0\)

Hence, option 2 is the correct answer.