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

Last updated on Apr 17, 2025

Latest Rigid Body Dynamics MCQ Objective Questions

Rigid Body Dynamics Question 1:

A stick of length l and mass M , initially upright on a frictionless table, starts falling under the influence of gravity. Using energy conservation, what is the expression for the speed ( \(\dot{y} \) ) of the center of mass as a function of the angle \(\theta\) from the vertical?

12-4-2025 IMG-680 -18

  1. \(\dot{y} = \sqrt{\frac{6g \sin^2 \theta}{3 \sin^2 \theta + 1}} \)
  2. \(\dot{y} = \frac{l}{2} \sin \theta \dot{\theta}\)
  3. \(\dot{y}^2 = \frac{2 g y}{[1 + (1/3) \sin^2 \theta]}\)
  4. \( \dot{y} = \sqrt{\frac{3 l g (1 - \cos \theta) \sin^2 \theta}{3 \sin^2 \theta + 1}} \)S

Answer (Detailed Solution Below)

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

Rigid Body Dynamics Question 1 Detailed Solution

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

 

Rigid Body Dynamics Question 2:

A uniform drum of radius b and mass M rolls without slipping down a plane inclined at an angle \( \theta \) . The moment of inertia of the drum about its axis is \( I_0 = \frac{1}{2} M b^2\) . What is the linear acceleration ( a ) of the drum along the plane?

 

  1. \( a = g \sin \theta\)
  2. \( a = \frac{2}{3} g \sin \theta\)
  3. \(a = \frac{3}{2} g \sin \theta\)
  4. \(a = \frac{g \sin \theta}{2}\)

Answer (Detailed Solution Below)

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

Rigid Body Dynamics Question 2 Detailed Solution

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

Rigid Body Dynamics Question 3:

A disk of mass M and radius b is pulled with a constant force F by a thin tape wound around its circumference, sliding on ice without friction. Which of the following is the angular acceleration ( \(\alpha\) ) of the disk?

 

  1. \(\alpha = \frac{2F}{mb}\)
  2. \(\alpha = \frac{F}{M} \)
  3. \( \alpha = \frac{4 F}{Mb} \)
  4. \(\alpha = 0\)

Answer (Detailed Solution Below)

Option 1 : \(\alpha = \frac{2F}{mb}\)

Rigid Body Dynamics Question 3 Detailed Solution

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:
     qImage673de4bde568204e8d44c1db
  \( \tau_0 = b F\)
     
      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 : 
     
    \(\alpha = \frac{2F}{mb}\)
    

Rigid Body Dynamics Question 4:

A uniform wheel of mass M and radius b rolls uniformly and without slipping. What is the total angular momentum (\( L_z \) ) of the wheel about the origin?

 

  1. \(L_z = \frac{1}{2} M b^2 \omega\)
  2. \(L_z = M b V\)
  3. \(L_z = -\frac{1}{2} M b^2 \omega\)
  4. \( L_z = -\frac{3}{2} M b^2 \omega\)

Answer (Detailed Solution Below)

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

Rigid Body Dynamics Question 4 Detailed Solution

 

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:
     
     \(L_z = -\frac{1}{2} M b^2 \omega - M b^2 \omega \)
     
\( L_z = -\frac{3}{2} M b^2 \omega\)
    

Rigid Body Dynamics Question 5:

A railroad crossing gate consists of a plank of mass M and length 2L , pivoted at one end. A support rod is used to minimize wear and tear on the pivot. At what distance l from the pivot should the support rod be placed to minimize the force on the pivot?

  1. \( l = \frac{L}{2}\)
  2. \(l = \frac{2L}{3}\)
  3. \( l = \frac{ML}{I_p}\)
  4. \(l = \frac{4L}{3}\)

Answer (Detailed Solution Below)

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

Rigid Body Dynamics Question 5 Detailed Solution

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.

qImage673de13f3003b62159d6e91a
    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:
     
   \( l = \frac{4L}{3}\)
     

Top Rigid Body Dynamics MCQ Objective Questions

Rigid Body Dynamics Question 6:

A cylindrical cavity of diameter 'a' exists inside a cylinder of diameter '2a' as shown in the figure. Both the cylinder and the cavity are infinitely long. A uniform current density \(J\) flows along the length. If the magnitude of the magnetic field at the point P is given by \(\frac{N}{12}\mu_{0}J\) , then the value of \(N\) is :
qImage671b299a78af04a8dc5907e4

  1. \(5\)
  2. \(6\)
  3. \(7\)
  4. \(8\)

Answer (Detailed Solution Below)

Option 1 : \(5\)

Rigid Body Dynamics Question 6 Detailed Solution

Calculation:
The magnetic field for an infinitely long cylinder is given by,

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

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

\( r = \) 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 = \( B = {B}_{1} + {B}_{2} \)

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

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

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

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

\( \therefore N = 5 \)

Rigid Body Dynamics Question 7:

A cylindrical rigid block has principal moments of inertia I about the symmetry axis and 2I about each of the perpendicular axes passing through the center of mass. At some instant, the components of angular momentum about the center of mass in the body-fixed principal axis frame is (l, l, l) with l > 0 . What is the cosine of the angle between the angular momentum and the angular velocity?

  1. \(\frac{2}{3}\)
  2. \(\frac{2}{\sqrt6}\)
  3. \(\frac{2\sqrt2}{3}\)
  4. \(\frac{5}{3\sqrt3}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{2\sqrt2}{3}\)

Rigid Body Dynamics Question 7 Detailed Solution

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

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

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

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

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

Rigid Body Dynamics Question 8:

A stick of length l and mass M , initially upright on a frictionless table, starts falling under the influence of gravity. Using energy conservation, what is the expression for the speed ( \(\dot{y} \) ) of the center of mass as a function of the angle \(\theta\) from the vertical?

12-4-2025 IMG-680 -18

  1. \(\dot{y} = \sqrt{\frac{6g \sin^2 \theta}{3 \sin^2 \theta + 1}} \)
  2. \(\dot{y} = \frac{l}{2} \sin \theta \dot{\theta}\)
  3. \(\dot{y}^2 = \frac{2 g y}{[1 + (1/3) \sin^2 \theta]}\)
  4. \( \dot{y} = \sqrt{\frac{3 l g (1 - \cos \theta) \sin^2 \theta}{3 \sin^2 \theta + 1}} \)S

Answer (Detailed Solution Below)

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

Rigid Body Dynamics Question 8 Detailed Solution

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

 

Rigid Body Dynamics Question 9:

A uniform drum of radius b and mass M rolls without slipping down a plane inclined at an angle \( \theta \) . The moment of inertia of the drum about its axis is \( I_0 = \frac{1}{2} M b^2\) . What is the linear acceleration ( a ) of the drum along the plane?

 

  1. \( a = g \sin \theta\)
  2. \( a = \frac{2}{3} g \sin \theta\)
  3. \(a = \frac{3}{2} g \sin \theta\)
  4. \(a = \frac{g \sin \theta}{2}\)

Answer (Detailed Solution Below)

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

Rigid Body Dynamics Question 9 Detailed Solution

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

Rigid Body Dynamics Question 10:

A disk of mass M and radius b is pulled with a constant force F by a thin tape wound around its circumference, sliding on ice without friction. Which of the following is the angular acceleration ( \(\alpha\) ) of the disk?

 

  1. \(\alpha = \frac{2F}{mb}\)
  2. \(\alpha = \frac{F}{M} \)
  3. \( \alpha = \frac{4 F}{Mb} \)
  4. \(\alpha = 0\)

Answer (Detailed Solution Below)

Option 1 : \(\alpha = \frac{2F}{mb}\)

Rigid Body Dynamics Question 10 Detailed Solution

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:
     qImage673de4bde568204e8d44c1db
  \( \tau_0 = b F\)
     
      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 : 
     
    \(\alpha = \frac{2F}{mb}\)
    

Rigid Body Dynamics Question 11:

A uniform wheel of mass M and radius b rolls uniformly and without slipping. What is the total angular momentum (\( L_z \) ) of the wheel about the origin?

 

  1. \(L_z = \frac{1}{2} M b^2 \omega\)
  2. \(L_z = M b V\)
  3. \(L_z = -\frac{1}{2} M b^2 \omega\)
  4. \( L_z = -\frac{3}{2} M b^2 \omega\)

Answer (Detailed Solution Below)

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

Rigid Body Dynamics Question 11 Detailed Solution

 

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:
     
     \(L_z = -\frac{1}{2} M b^2 \omega - M b^2 \omega \)
     
\( L_z = -\frac{3}{2} M b^2 \omega\)
    

Rigid Body Dynamics Question 12:

A railroad crossing gate consists of a plank of mass M and length 2L , pivoted at one end. A support rod is used to minimize wear and tear on the pivot. At what distance l from the pivot should the support rod be placed to minimize the force on the pivot?

  1. \( l = \frac{L}{2}\)
  2. \(l = \frac{2L}{3}\)
  3. \( l = \frac{ML}{I_p}\)
  4. \(l = \frac{4L}{3}\)

Answer (Detailed Solution Below)

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

Rigid Body Dynamics Question 12 Detailed Solution

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.

qImage673de13f3003b62159d6e91a
    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:
     
   \( l = \frac{4L}{3}\)
     

Rigid Body Dynamics Question 13:

Consider a conical pendulum consisting of a bob of mass M attached to a string, which rotates uniformly with a constant angular velocity. If we take the origin at the point where the string is fixed and assume a uniform circular motion, what is the torque ( \(\tau_B\) ) about a point at the center of the circular path?

12-4-2025 IMG-680 -19

  1. \(\tau_B = 0\)
  2. \(\tau_B = M g l \sin \alpha \hat{\theta} \)
  3. \(\tau_B = M l v^2 \cos \alpha \hat{k}\)
  4. \(\tau_B = M r \omega^2 \hat{\theta}\)

Answer (Detailed Solution Below)

Option 2 : \(\tau_B = M g l \sin \alpha \hat{\theta} \)

Rigid Body Dynamics Question 13 Detailed Solution

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:

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

Rigid Body Dynamics Question 14:

 A block of mass m slides along the x axis with velocity \( \mathbf{v} = v \hat{i} \) . If the block encounters a friction force \(\mathbf{f} = -f \hat{i} \) , what is the time rate of change of angular momentum ( \( d\mathbf{L}_B/dt \) ) about point B ?

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  1. \( \frac{d\mathbf{L}_B}{dt} = 0\)
  2. \(\frac{d\mathbf{L}_B}{dt} = -f l \hat{k}\)
  3. \(\frac{d\mathbf{L}_B}{dt} = -l v \hat{k}\)
  4. \(\frac{d\mathbf{L}_B}{dt} = f l \hat{i} \)

Answer (Detailed Solution Below)

Option 2 : \(\frac{d\mathbf{L}_B}{dt} = -f l \hat{k}\)

Rigid Body Dynamics Question 14 Detailed Solution

 

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, rB 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:

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

  

Rigid Body Dynamics Question 15:

A spacecraft is aimed at a far off planet with radius R . Due to the gravitational influence of the planet, the effective area ( A e ) for hitting the planet is greater than its geometrical area ( A g ). What is the effective area A e of the capture cross section in terms of the geometrical area?

 

  1. \(A_e = A_g\)
  2. \(A_e = A_g \left( 1 + \frac{mMG/R}{mv_0^2/2} \right) \\\)
  3. \(A_e = A_g \left( 1 -\frac{mMG/R}{mv_0^2/2} \right) \\\)
  4. \( A_e = \pi R^2\)

Answer (Detailed Solution Below)

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

Rigid Body Dynamics Question 15 Detailed Solution

 

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.

 

 

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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\( b'^2 \) :

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

 

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