"Rigid Body Dynamics, Newton’s Law of Motion, Circular Motion, Center of Mass Online MCQ Test "
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1. The moment of inertia of a door of mass \( m \), length \( 2\ell \), and width \( \ell \) about its longer side is:
A. \( \frac{24}{5} m\ell^2 \)
B. \( \frac{5}{24} m\ell^2 \)
C. \( \frac{1}{3} m\ell^2 \)
D. None of these
The correct answer is C. \( \frac{1}{3} m\ell^2 \).
Step-by-Step Solution:
1. Moment of Inertia Formula:
For a rectangular object (such as a door) rotated about one of its edges, the moment of inertia is calculated using: \[ I = \frac{1}{3} mL^2 \] where \( L \) is the length of the door, and \( m \) is its mass.
Conclusion:
Thus, the moment of inertia of the door about its longer side is \( \frac{1}{3} m\ell^2 \).
Step-by-Step Solution:
1. Moment of Inertia Formula:
For a rectangular object (such as a door) rotated about one of its edges, the moment of inertia is calculated using: \[ I = \frac{1}{3} mL^2 \] where \( L \) is the length of the door, and \( m \) is its mass.
Thus, the moment of inertia of the door about its longer side is \( \frac{1}{3} m\ell^2 \).
2. Two blocks of equal mass are tied with a light string which passes over a massless pulley as shown in the figure. The magnitude of the acceleration of the center of mass of both blocks is (neglect friction everywhere):
A. \( \frac{\sqrt{3}-1g}{4\sqrt2} \)
B. \( (3 - 1)g \)
C. \( \frac{2g}{3} \)
D. \( \frac{13g}{2} \)
The correct answer is A. \( \frac{\sqrt{3}-1g}{4\sqrt2} \).
3. Three point masses are arranged as shown in the figure. The moment of inertia of the system about the axis \( OO' \) (passing through its plane) is:
A. \( 2m a^2 \)
B. \( m\frac{a^2}{2} \)
C. \( m a^2 \)
D. None of these
The correct answer is B. \( m\frac{a^2}{2} \).
Step-by-Step Solution:
1. Understanding the Moment of Inertia:
Moment of inertia about an axis is calculated using the formula: \[ I = \sum m_i r_i^2 \] where \( m_i \) is the mass of the particle and \( r_i \) is its perpendicular distance from the axis of rotation.
2. Analysis of the Given System:
- Three masses are arranged in the plane.
- The distances of the masses from the axis \( OO' \) are given as \( r_1 =\frac{ a}{2} \), \( r_2 = \frac{ a}{2} \), and \( r_3 = 0 \).
- Mass of each particle is \( m \).
3. Calculating the Moment of Inertia:
Using the formula: \[ I = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 \] Substituting the values: \[ I = m\frac{a^2}{2^2} + m\frac{a^2}{2^2} + m(0^2) \] \[ I = m\frac{a^2}{4} + m\frac{a^2}{4} = m\frac{a^2}{2} \] \[ I = m\frac{a^2}{2} \]
4. Conclusion:
Thus, the moment of inertia of the system about the axis \( OO' \) is \(m\frac{a^2}{2} \), which corresponds to option B.
Step-by-Step Solution:
1. Understanding the Moment of Inertia:
Moment of inertia about an axis is calculated using the formula: \[ I = \sum m_i r_i^2 \] where \( m_i \) is the mass of the particle and \( r_i \) is its perpendicular distance from the axis of rotation.
2. Analysis of the Given System:
- Three masses are arranged in the plane.
- The distances of the masses from the axis \( OO' \) are given as \( r_1 =\frac{ a}{2} \), \( r_2 = \frac{ a}{2} \), and \( r_3 = 0 \).
- Mass of each particle is \( m \).
3. Calculating the Moment of Inertia:
Using the formula: \[ I = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 \] Substituting the values: \[ I = m\frac{a^2}{2^2} + m\frac{a^2}{2^2} + m(0^2) \] \[ I = m\frac{a^2}{4} + m\frac{a^2}{4} = m\frac{a^2}{2} \] \[ I = m\frac{a^2}{2} \]
4. Conclusion:
Thus, the moment of inertia of the system about the axis \( OO' \) is \(m\frac{a^2}{2} \), which corresponds to option B.
4. A section of a fixed smooth circular track of radius \( R \) in the vertical plane is shown in the figure. A block is released from position \( A \) and leaves the track at \( B \). The radius of curvature of its trajectory when it just leaves the track at \( B \) is:
A. \( R \)
B. \( \frac{R}{4} \)
C. \( \frac{R}{2} \)
D. None of these
The correct answer is C. \( \frac{R}{2} \).
Step-by-Step Solution:
1. Energy Conservation:
When the block is released from rest at position \( A \), it has potential energy but no kinetic energy. As it moves down the track and reaches \( B \), its total mechanical energy is conserved: \[ \text{At A: } \text{Total Energy } = m g h = m g (2R) \] \[ \text{At B: } \text{Total Energy } = \text{Kinetic Energy} + \text{Potential Energy} \] Substituting the height at \( B \), \( h = R \): \[ m g \frac{2R}{5} + 0 = \frac{1}{2} m v^2 + \frac{m g R}{5} \] Simplify: \[ \frac{m g}{5} (2R - R) = \frac{1}{2} m v^2 \implies \frac{m g R}{5} = \frac{1}{2} m v^2 \] \[ v^2 = \frac{2gR}{5}\]
2. Radius of Curvature:
The radius of curvature \( r \) at the point of leaving the track is given by: \[ r = \frac{v^2}{g \cos \theta} \] At \( B \), the angle \( \theta = 37^\circ \), so \( \cos \theta = \frac{4}{5} \): \[ r = \frac{v^2}{g \cos 37^\circ} = \frac{2gR}{5}{g \cdot \frac{4}{5}} \] Simplify: \[ r = \frac{2R}{\frac{4}{5}} = \frac{2R \cdot 5}{4} = \frac{10R}{4} = \frac{R}{2} \]
3. Conclusion:
The radius of curvature of the trajectory when the block just leaves the track is \( \frac{R}{2} \), which matches option C.
Step-by-Step Solution:
1. Energy Conservation:
When the block is released from rest at position \( A \), it has potential energy but no kinetic energy. As it moves down the track and reaches \( B \), its total mechanical energy is conserved: \[ \text{At A: } \text{Total Energy } = m g h = m g (2R) \] \[ \text{At B: } \text{Total Energy } = \text{Kinetic Energy} + \text{Potential Energy} \] Substituting the height at \( B \), \( h = R \): \[ m g \frac{2R}{5} + 0 = \frac{1}{2} m v^2 + \frac{m g R}{5} \] Simplify: \[ \frac{m g}{5} (2R - R) = \frac{1}{2} m v^2 \implies \frac{m g R}{5} = \frac{1}{2} m v^2 \] \[ v^2 = \frac{2gR}{5}\]
2. Radius of Curvature:
The radius of curvature \( r \) at the point of leaving the track is given by: \[ r = \frac{v^2}{g \cos \theta} \] At \( B \), the angle \( \theta = 37^\circ \), so \( \cos \theta = \frac{4}{5} \): \[ r = \frac{v^2}{g \cos 37^\circ} = \frac{2gR}{5}{g \cdot \frac{4}{5}} \] Simplify: \[ r = \frac{2R}{\frac{4}{5}} = \frac{2R \cdot 5}{4} = \frac{10R}{4} = \frac{R}{2} \]
3. Conclusion:
The radius of curvature of the trajectory when the block just leaves the track is \( \frac{R}{2} \), which matches option C.
5. In the figure, the block \( B \) of mass \( m \) starts from rest at the top of a wedge \( W \) of mass \( M \). All surfaces are without friction. \( W \) can slide on the ground. \( B \) slides down onto the ground, moves along the ground with speed \( v \), has an elastic collision with the wall, and climbs back onto \( W \).
A. \( B \) will reach the top of \( W \) again.
B. From the beginning, till the collision with the wall, the center of mass of 'B + W' is stationary in the horizontal direction.
C. After the collision, the center of mass of 'B + W' moves with the velocity \( \frac{2mv}{M + m} \).
D. When \( B \) reaches its highest position on \( W \), the speed of \( W \) is \( \frac{2mv}{M + m} \).
The correct answers are B, C, and D.
Step-by-Step Solution:
1. Conservation of Momentum:
Initially, the system is at rest, and since there are no external horizontal forces acting, the horizontal momentum of the system remains conserved throughout.
\[ \text{Momentum of system: } m v = M V \] Therefore, the velocity of \( W \) is: \[ V = \frac{mv}{M} \]
2. Motion of the Center of Mass:
Since the total momentum of the system is conserved, the center of mass of the combined system (block \( B \) and wedge \( W \)) does not move in the horizontal direction until the collision with the wall.
Conclusion:
- Statement B is true because the center of mass remains stationary until the collision.
- Statement C is true because after the collision, the center of mass velocity is \( \frac{2mv}{M + m} \).
- Statement D is true because the speed of \( W \) when \( B \) reaches its highest position is \( \frac{2mv}{M + m} \).
Step-by-Step Solution:
1. Conservation of Momentum:
Initially, the system is at rest, and since there are no external horizontal forces acting, the horizontal momentum of the system remains conserved throughout.
\[ \text{Momentum of system: } m v = M V \] Therefore, the velocity of \( W \) is: \[ V = \frac{mv}{M} \]
2. Motion of the Center of Mass:
Since the total momentum of the system is conserved, the center of mass of the combined system (block \( B \) and wedge \( W \)) does not move in the horizontal direction until the collision with the wall.
- Statement B is true because the center of mass remains stationary until the collision.
- Statement C is true because after the collision, the center of mass velocity is \( \frac{2mv}{M + m} \).
- Statement D is true because the speed of \( W \) when \( B \) reaches its highest position is \( \frac{2mv}{M + m} \).
6. In a free space, a rifle of mass \( M \) shoots a bullet of mass \( m \) at a stationary block of mass \( M \), a distance \( D \) away from it. When the bullet has moved through a distance \( d \) towards the block, the center of mass of the bullet-block system is at a distance of:
A. \( \frac{(D - d)m}{M + m} \) from the block
B. \( \frac{md + MD}{M + m} \) from the rifle
C. \( \frac{2dm + DM}{M + m} \) from the rifle
D. \( (D - d) \frac{M}{M + m} \) from the bullet
The correct answers are A and D.
Step-by-Step Solution:
1. Understanding the System:
- The rifle shoots a bullet of mass \( m \) with speed, causing the bullet to travel a distance \( d \).
- A stationary block of mass \( M \) is placed a distance \( D \) from the rifle.
- The system consists of the bullet and the block, and the center of mass position depends on their masses and positions.
2. Position of the Center of Mass:
The center of mass position \( x_{\text{cm}} \) of a system is calculated as: \[ x_{\text{cm}} = \frac{m x_1 + M x_2}{M + m} \] where: - \( x_1 = \text{distance of the bullet from the block} = D - d \), - \( x_2 = 0 \) (the block is stationary).
Substituting values: \[ x_{\text{cm}} = \frac{m(D - d) + M(0)}{M + m} = \frac{(D - d)m}{M + m} \] Thus, the center of mass is at a distance \( \frac{(D - d)m}{M + m} \) from the block.
3. Relative Position from the Bullet:
The center of mass relative to the bullet is calculated similarly: \[ x_{\text{cm}} = (D - d) \frac{M}{M + m} \] Thus, the center of mass is also \( (D - d) \frac{M}{M + m} \) from the bullet.
4. Conclusion:
- Option A is correct because the center of mass is \( \frac{(D - d)m}{M + m} \) from the block.
- Option D is correct because the center of mass is \( (D - d) \frac{M}{M + m} \) from the bullet.
Step-by-Step Solution:
1. Understanding the System:
- The rifle shoots a bullet of mass \( m \) with speed, causing the bullet to travel a distance \( d \).
- A stationary block of mass \( M \) is placed a distance \( D \) from the rifle.
- The system consists of the bullet and the block, and the center of mass position depends on their masses and positions.
2. Position of the Center of Mass:
The center of mass position \( x_{\text{cm}} \) of a system is calculated as: \[ x_{\text{cm}} = \frac{m x_1 + M x_2}{M + m} \] where: - \( x_1 = \text{distance of the bullet from the block} = D - d \), - \( x_2 = 0 \) (the block is stationary).
Substituting values: \[ x_{\text{cm}} = \frac{m(D - d) + M(0)}{M + m} = \frac{(D - d)m}{M + m} \] Thus, the center of mass is at a distance \( \frac{(D - d)m}{M + m} \) from the block.
3. Relative Position from the Bullet:
The center of mass relative to the bullet is calculated similarly: \[ x_{\text{cm}} = (D - d) \frac{M}{M + m} \] Thus, the center of mass is also \( (D - d) \frac{M}{M + m} \) from the bullet.
4. Conclusion:
- Option A is correct because the center of mass is \( \frac{(D - d)m}{M + m} \) from the block.
- Option D is correct because the center of mass is \( (D - d) \frac{M}{M + m} \) from the bullet.
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