Rigid Body Dynamics MCQ Quiz - Objective Question with Answer for Rigid Body Dynamics - Download Free PDF

Correct Option : 4)\( \dot{y} = \sqrt{\frac{3 l g (1 - \cos \theta) \sin^2 \theta}{3 \sin^2 \theta + 1}}\)

Explanation :
    The initial potential energy (\( U_0 \) ) when the stick is upright is:
   \( U_0 = M g \frac{l}{2}\)
     
    Since the stick starts from rest, the initial kinetic energy ( \(K_0\) ) is zero, so the total initial energy is:
     
  \( E = K_0 + U_0 = M g \frac{l}{2} \)

    As the stick falls, it rotates and the center of mass falls vertically down a distance y . The kinetic energy ( K ) of the system at an angle \theta includes both rotational and translational parts:
     
  \( K = \frac{1}{2} I_0 \dot{\theta}^2 + \frac{1}{2} M \dot{y}^2 \)
    The potential energy at a distance y is:
     
     \(U = M g \left( \frac{l}{2} - y \right) \)

    Since mechanical energy is conserved:
     
   \( K + U = K_0 + U_0 = M g \frac{l}{2}\)
     
    Substituting the expressions for K and U :
     
   \( \frac{1}{2} M \dot{y}^2 + \frac{1}{2} I_0 \dot{\theta}^2 + M g \left( \frac{l}{2} - y \right) = M g \frac{l}{2} \)

    The relation between y and \theta is given by:
     
     \(y = \frac{l}{2} (1 - \cos \theta)\)
     
    Differentiating y with respect to time gives:
\( \dot{y} = \frac{l}{2} \sin \theta \dot{\theta}\)
     
    Substitute the expression for\( I_0 = \frac{M l^2}{12}\) and eliminate \( \dot{\theta}\) :
     
     \(\frac{1}{2} M \dot{y}^2 + \frac{1}{2} \frac{M l^2}{12} \left( \frac{2 \dot{y}}{l \sin \theta} \right)^2 + M g \left( \frac{l}{2} - y \right) = M g \frac{l}{2} \)
    Rearranging and simplifying yields:
     
    \( \dot{y} = \sqrt{\frac{3 l g (1 - \cos \theta) \sin^2 \theta}{3 \sin^2 \theta + 1}} \)

Correct Option : 2) \(a = \frac{2}{3} g \sin \theta\)

Explanation :

1. Method 1: Translational and Rotational Dynamics :
    The forces acting on the drum are gravity, friction, and the normal force.
    The equation of motion for translation of the center of mass along the plane is:
     
\( W \sin \theta - f = M a\)
     
     where W = Mg and f is the frictional force.
    For rotation around the center of mass:
     
\( b f = I_0 \alpha\)
     
     Since the drum rolls without slipping, we have:
     
 \( a = b \alpha\)
     

2. Eliminating Friction :
    Substitute f from the rotational equation into the translational equation:
     
    \( M g \sin \theta - \frac{I_0}{b} \alpha = M a\)
     
    Using \(I_0 = \frac{1}{2} M b^2 \text{ and } \alpha = \frac{a}{b}\) :
     
     \(M g \sin \theta - \frac{M b^2}{2b} \cdot \frac{a}{b} = M a\)
     
    Simplify:
     
   \( M g \sin \theta - \frac{M a}{2} = M a\)
     
    Rearranging to solve for a :
     
     \(a = \frac{2}{3} g \sin \theta\)
     

3. Method 2: Using Torque About Point A :
    Consider a coordinate system with the origin at point A , the point of contact between the drum and the plane.
    The torque around point A is:
     
  \( \tau_z = (R \times F)_z = -b W \sin \theta \)
    Using the angular momentum equation and applying \(\tau_z = \frac{dL_z}{dt}\) , we find:
     
     \(a = \frac{2}{3} g \sin \theta\)
     

Correct Option : A) \(\alpha = \frac{2F}{mb}\)

Explanation :
    The disk is acted upon by a force F at its circumference, causing a torque ( \(\tau_0\) ) about its center of mass:
     
  \( \tau_0 = b F\)
     
      Where b is the radius of the disk.
    Using the relation between torque and angular acceleration (\( α\) ):
     
  \( \tau_0 = \frac{1}{2}Mb^2α \)
     
     \(b F = \frac{1}{2}Mb^2α \)
     
    Solving for α : 
     
    \(\alpha = \frac{2F}{mb}\)
    

Correct Option : 4) \(L_z = -\frac{3}{2} M b^2 \omega\)

Explanation :

Angular Momentum About the Center of Mass :
    The moment of inertia of the wheel about its center of mass is given by:
     
    \( I_0 = \frac{1}{2} M b^2\)
     
    Therefore, the angular momentum about the center of mass is:
     
    \( L_0 = -I_0 \omega = -\frac{1}{2} M b^2 \omega\)
     
     The negative sign indicates that the direction of \(L_0\) is into the paper, along the negative z axis.

 Rolling Without Slipping :
    Since the wheel rolls without slipping, the velocity V of the center of mass is related to the angular velocity by:
     
    \( V = b \omega\)
     
    The contribution of the linear motion to the angular momentum about the origin is given by:
     
    \( (R \times MV)_z = -M b V = -M b^2 \omega\)
     

Total Angular Momentum About the Origin :
    The total angular momentum about the origin ( \(L_z \) ) is the sum of the angular momentum due to rotation about the center of mass and the contribution from the linear motion:
     
    \( L_z = L_0 + (R \times MV)_z \)
     Substituting the values:
     
     \(L_z = -\frac{1}{2} M b^2 \omega - M b^2 \omega \)
     
\( L_z = -\frac{3}{2} M b^2 \omega\)
    

Correct Option : 4) \(l = \frac{4L}{3}\)

Explanation :
    Taking torques around the pivot, we have:
     
    \( \tau = MgL - F_{sv} l = I_p \ddot{\theta}\)
     
     where \(I_p\) is the moment of inertia about the pivot, L is the distance from the pivot to the center of mass, and l is the distance from the pivot to the support rod.

    To minimize the force \(F_{pv}\) due to the pivot during the short collision time, we need the impulse contribution from \(F_{pv}\) to vanish.
    By solving the equations derived from torque and impulse considerations, it turns out that minimizing the force on the pivot requires setting:
     
 \( F_{pv} \Delta t = \left( \frac{I_p}{l} - ML \right) \dot{\theta} \)
     Setting \( F_{pv} = 0\) leads to:
    \( \frac{I_p}{l} - ML = 0\)
     

3. Optimal Placement of Support Rod :
    Rearranging to solve for l :
     
    \( l = \frac{I_p}{ML}\)
     
    Assuming the gate is like a long, thin rod pivoted at one end, the moment of inertia\( I_p\) is given by:
     
    \( I_p = \frac{M(2L)^2}{3} = \frac{4ML^2}{3}\)
     
     Substituting back gives:
     
   \( l = \frac{4L}{3}\)
     

\( { B }_{ in } = \dfrac { { \mu }_{ 0 }Jr }{ 2 } \)

\( { B }_{ out } = \dfrac { { \mu }_{ 0 }J{R}^{2} }{ 2 r } \)

\( r = \) distance from the axis of the cylinder.

\( R = \) Radius of the cylinder.

Assuming the bigger cylinder to carry a positive current density and the smaller cylinder carry a negative current density of magnitude J each.

\( \therefore \) Magnetic field at point P = \( B = {B}_{1} + {B}_{2} \)

\( { B }_{ 1 } = \dfrac { { \mu }_{ 0 }Ja }{ 2 } \)

\( { B }_{ 2 } = \dfrac { -{ \mu }_{ 0 }J{ (\dfrac { a }{ 2 } ) }^{ 2 } }{ 2\dfrac { 3a }{ 2 } } \)

\( \therefore { B }_{ 2 } = \dfrac {- { \mu }_{ 0 }Ja }{ 12 } \)

\( \therefore { B } = \dfrac { { 5\mu }_{ 0 }Ja }{ 12 } \)

\( \therefore N = 5 \)

\(\rm \vec{L}=\overline{\bar{I}} \vec{\omega} \)

\(\left(\begin{array}{l} l \\ l \\ l \end{array}\right)=\left(\begin{array}{ccc} 2 I & 0 & 0 \\ 0 & 2 I & 0 \\ 0 & 0 & I \end{array}\right)\left(\begin{array}{l} \omega_{x} \\ \omega_{y} \\ \omega_{z} \end{array}\right) \Rightarrow \omega_{x}=\frac{l}{2 I}, \omega_{y}=\frac{l}{2 I}, \omega_{z}=\frac{l}{I}\)

\(\rm \vec{L}=l \hat{x}+l \hat{y}+l \hat{z}, \vec{\omega}=\frac{l}{2 I} \hat{x}+\frac{l}{2 I} \hat{y}+\frac{l}{I} \hat{z}\)

\(\rm \cos \theta=\frac{\vec{L} \cdot \vec{\omega}}{L \omega}\) = \(\rm \frac{\frac{l^2}{2I}+\frac{l^2}{2I}+\frac{l^2}{I}}{\sqrt{l^{2}+l^{2}+l^{2}} \sqrt{\frac{l^2}{4I^2}+ \frac{l^2}{4I^2}+\frac{l^2}{2I^2}}}\) = \(\rm \frac{\frac{4l^2}{2I}}{l \sqrt{3} \sqrt{\frac{6l^2}{4I^2}}}=\frac{\frac{4l^2}{2I}}{\frac{l^2}{2I}\sqrt{18}}=\frac{4}{\sqrt{18}}=\frac{4}{3 \sqrt{2}} \)

\(\Rightarrow \cos \theta=\frac{2 \sqrt{2}}{3}\)

Correct Option : 4)\( \dot{y} = \sqrt{\frac{3 l g (1 - \cos \theta) \sin^2 \theta}{3 \sin^2 \theta + 1}}\)

Explanation :
    The initial potential energy (\( U_0 \) ) when the stick is upright is:
   \( U_0 = M g \frac{l}{2}\)
     
    Since the stick starts from rest, the initial kinetic energy ( \(K_0\) ) is zero, so the total initial energy is:
     
  \( E = K_0 + U_0 = M g \frac{l}{2} \)

    As the stick falls, it rotates and the center of mass falls vertically down a distance y . The kinetic energy ( K ) of the system at an angle \theta includes both rotational and translational parts:
     
  \( K = \frac{1}{2} I_0 \dot{\theta}^2 + \frac{1}{2} M \dot{y}^2 \)
    The potential energy at a distance y is:
     
     \(U = M g \left( \frac{l}{2} - y \right) \)

    Since mechanical energy is conserved:
     
   \( K + U = K_0 + U_0 = M g \frac{l}{2}\)
     
    Substituting the expressions for K and U :
     
   \( \frac{1}{2} M \dot{y}^2 + \frac{1}{2} I_0 \dot{\theta}^2 + M g \left( \frac{l}{2} - y \right) = M g \frac{l}{2} \)

    The relation between y and \theta is given by:
     
     \(y = \frac{l}{2} (1 - \cos \theta)\)
     
    Differentiating y with respect to time gives:
\( \dot{y} = \frac{l}{2} \sin \theta \dot{\theta}\)
     
    Substitute the expression for\( I_0 = \frac{M l^2}{12}\) and eliminate \( \dot{\theta}\) :
     
     \(\frac{1}{2} M \dot{y}^2 + \frac{1}{2} \frac{M l^2}{12} \left( \frac{2 \dot{y}}{l \sin \theta} \right)^2 + M g \left( \frac{l}{2} - y \right) = M g \frac{l}{2} \)
    Rearranging and simplifying yields:
     
    \( \dot{y} = \sqrt{\frac{3 l g (1 - \cos \theta) \sin^2 \theta}{3 \sin^2 \theta + 1}} \)

Correct Option : 2) \(a = \frac{2}{3} g \sin \theta\)

Explanation :

1. Method 1: Translational and Rotational Dynamics :
    The forces acting on the drum are gravity, friction, and the normal force.
    The equation of motion for translation of the center of mass along the plane is:
     
\( W \sin \theta - f = M a\)
     
     where W = Mg and f is the frictional force.
    For rotation around the center of mass:
     
\( b f = I_0 \alpha\)
     
     Since the drum rolls without slipping, we have:
     
 \( a = b \alpha\)
     

2. Eliminating Friction :
    Substitute f from the rotational equation into the translational equation:
     
    \( M g \sin \theta - \frac{I_0}{b} \alpha = M a\)
     
    Using \(I_0 = \frac{1}{2} M b^2 \text{ and } \alpha = \frac{a}{b}\) :
     
     \(M g \sin \theta - \frac{M b^2}{2b} \cdot \frac{a}{b} = M a\)
     
    Simplify:
     
   \( M g \sin \theta - \frac{M a}{2} = M a\)
     
    Rearranging to solve for a :
     
     \(a = \frac{2}{3} g \sin \theta\)
     

3. Method 2: Using Torque About Point A :
    Consider a coordinate system with the origin at point A , the point of contact between the drum and the plane.
    The torque around point A is:
     
  \( \tau_z = (R \times F)_z = -b W \sin \theta \)
    Using the angular momentum equation and applying \(\tau_z = \frac{dL_z}{dt}\) , we find:
     
     \(a = \frac{2}{3} g \sin \theta\)
     

Correct Option : A) \(\alpha = \frac{2F}{mb}\)

Explanation :
    The disk is acted upon by a force F at its circumference, causing a torque ( \(\tau_0\) ) about its center of mass:
     
  \( \tau_0 = b F\)
     
      Where b is the radius of the disk.
    Using the relation between torque and angular acceleration (\( α\) ):
     
  \( \tau_0 = \frac{1}{2}Mb^2α \)
     
     \(b F = \frac{1}{2}Mb^2α \)
     
    Solving for α : 
     
    \(\alpha = \frac{2F}{mb}\)
    

Correct Option : 4) \(L_z = -\frac{3}{2} M b^2 \omega\)

Explanation :

Angular Momentum About the Center of Mass :
    The moment of inertia of the wheel about its center of mass is given by:
     
    \( I_0 = \frac{1}{2} M b^2\)
     
    Therefore, the angular momentum about the center of mass is:
     
    \( L_0 = -I_0 \omega = -\frac{1}{2} M b^2 \omega\)
     
     The negative sign indicates that the direction of \(L_0\) is into the paper, along the negative z axis.

 Rolling Without Slipping :
    Since the wheel rolls without slipping, the velocity V of the center of mass is related to the angular velocity by:
     
    \( V = b \omega\)
     
    The contribution of the linear motion to the angular momentum about the origin is given by:
     
    \( (R \times MV)_z = -M b V = -M b^2 \omega\)
     

Total Angular Momentum About the Origin :
    The total angular momentum about the origin ( \(L_z \) ) is the sum of the angular momentum due to rotation about the center of mass and the contribution from the linear motion:
     
    \( L_z = L_0 + (R \times MV)_z \)
     Substituting the values:
     
     \(L_z = -\frac{1}{2} M b^2 \omega - M b^2 \omega \)
     
\( L_z = -\frac{3}{2} M b^2 \omega\)
    

Correct Option : 4) \(l = \frac{4L}{3}\)

Explanation :
    Taking torques around the pivot, we have:
     
    \( \tau = MgL - F_{sv} l = I_p \ddot{\theta}\)
     
     where \(I_p\) is the moment of inertia about the pivot, L is the distance from the pivot to the center of mass, and l is the distance from the pivot to the support rod.

    To minimize the force \(F_{pv}\) due to the pivot during the short collision time, we need the impulse contribution from \(F_{pv}\) to vanish.
    By solving the equations derived from torque and impulse considerations, it turns out that minimizing the force on the pivot requires setting:
     
 \( F_{pv} \Delta t = \left( \frac{I_p}{l} - ML \right) \dot{\theta} \)
     Setting \( F_{pv} = 0\) leads to:
    \( \frac{I_p}{l} - ML = 0\)
     

3. Optimal Placement of Support Rod :
    Rearranging to solve for l :
     
    \( l = \frac{I_p}{ML}\)
     
    Assuming the gate is like a long, thin rod pivoted at one end, the moment of inertia\( I_p\) is given by:
     
    \( I_p = \frac{M(2L)^2}{3} = \frac{4ML^2}{3}\)
     
     Substituting back gives:
     
   \( l = \frac{4L}{3}\)
     

Correct Option : B) \(\tau_B = M g l \sin \alpha \hat{\theta} \)

Explanation :

    For a conical pendulum, if we consider the torque about a point at the center of the circular path, the net force acting on the bob is radially inward:

         \( \mathbf{F} = - T \sin \alpha \hat{r} \)

    The torque ( \(\tau_B\) ) at point B , which is the center of the circular path, is given by:

     \(\tau_B = \mathbf{r}_B \times \mathbf{F}\)

    Calculating the magnitude of \( \tau_B \) :

        \( |\tau_B| = l F \cos \alpha = l T \cos \alpha \sin \alpha = M g l \sin \alpha \)

    The direction of torque ( \( \tau_B \)) is tangential to the line of motion of the mass M . Thus:

        \( \tau_B = M g l \sin \alpha \hat{\theta} \)

Correct Option : B) \(\frac{d\mathbf{L}_B}{dt} = -f l \hat{k} \)

Explanation :

    The angular momentum ( \( L_B\) ) of the block about point B can be expressed as:

   \( L_B = m r_B \times \mathbf{v} = m v l \hat{k}\)

     Here, r_B is the position vector from point B to the block, and l is the perpendicular distance from B to the line of motion of the block.

      The initial angular momentum ( \( L_B \)) is in the positive z direction ( \(\hat{k} \) ).

    When a friction force \(\mathbf{f} = -f \hat{i}\) acts on the block, a torque ( \(\tau_B\) ) is generated about point B :

           \(\tau_B = r_B \times \mathbf{f} = - l f \hat{k} \)

     The negative sign indicates that the torque is in the negative z direction.

    By Newton's Second Law of Rotational Motion:

\( \frac{d\mathbf{L}_B}{dt} = \tau_B\)

    Substituting the value of torque:

           \(\frac{d\mathbf{L}_B}{dt} = -f l \hat{k} \)

Correct Option : C) \(A_e = A_g \left( 1 + \frac{mMG/R}{mv_0^2/2} \right) \\\)

Explanation :

    The effective area for capturing the spacecraft is influenced by the gravitational attraction of the planet, leading to an increase in the area of impact compared to the geometrical area.

    The effective area is given by:

     \(A_e = \pi (b')^2\),

     where b' is the impact parameter that takes into account the gravitational influence.

Angular momentum L is conserved because there are no external torques acting on the spacecraft about the center of the planet.

Initial Angular Momentum :
 
 \( L_i = -m b' v_0 \) 
  where m is the mass of the spacecraft, b' is the impact parameter, and \(v_0\) is the initial velocity.

Angular Momentum at Closest Approach :
  At the point of closest approach, the radial distance is r = R , and velocity is v(R) perpendicular to r :
\( L_c = -m R v(R)\)
 
  
By conservation of angular momentum:
 
  \(-m b' v_0 = -m R v(R)\)
 
  Solving for v(R) :
 
\( v(R) = v_0 \frac{b'}{R}\)
 
Total mechanical energy E is conserved.

Initial Total Energy :
 
 \( E_i = \frac{1}{2} m v_0^2\)
 

Total Energy at Closest Approach :
  At the point of closest approach r = R :
  Kinetic Energy :
   
    \(K_c = \frac{1}{2} m v(R)^2\)
   
  Potential Energy :
   
    \(U_c = -\frac{mMG}{R}\)
   
    where M is the mass of the planet and G is the gravitational constant.

  Total Energy at Closest Approach :
   
  \( E_c = \frac{1}{2} m v(R)^2 - \frac{mMG}{R}\)
   

By conservation of energy:
 
 \( \frac{1}{2} m v_0^2 = \frac{1}{2} m v(R)^2 - \frac{mMG}{R}\)
 
  Substitute \(v(R) = v_0 \frac{b'}{R}\) :
 
 \( \frac{1}{2} m v_0^2 = \frac{1}{2} m \left( v_0 \frac{b'}{R} \right)^2 - \frac{mMG}{R}\)
 

  Simplify and solve for b' :
 
\( \frac{1}{2} m v_0^2 = \frac{1}{2} m \frac{v_0^2 b'^2}{R^2} - \frac{mMG}{R} \)

  Divide by m and multiply everything by 2:
 
 \( v_0^2 = \frac{v_0^2 b'^2}{R^2} - \frac{2MG}{R} \)

  Rearrange to solve for b'^2 :
 
\( b'^2 = R^2 \left( 1 + \frac{2MG}{v_0^2 R} \right)\)
 
The effective area for hitting the planet is given by:
\( A_e = \pi (b')^2\)

Substituting the value of\( b'^2 \) :

\(A_e = \pi R^2 \left( 1 + \frac{2MG}{v_0^2 R} \right)\)