System of particles and Rotational motion Important Questions
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1. A particle performing uniform circular motion has angular momentum \( L \). If its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is:
A. \( \frac{L}{2} \)
B. \( \frac{L}{4} \)
C. \( 2L \)
D. \( 4L \)
The correct answer is (b) \( L/4 \).
Solution:
1. Initial Conditions: - Let the initial angular momentum of the particle be \( L \). - The angular frequency (ω) of rotation is given by: \[ \omega = \frac{2\pi}{T} \] where \( T \) is the period of rotation. 2. Angular Momentum: - Angular momentum \( L \) is given by: \[ L = I\omega \] where \( I \) is the moment of inertia of the particle. - The kinetic energy \( K \) of the rotating particle is: \[ K = \frac{1}{2} I \omega^2 \] 3. Changes in Angular Frequency: - If the angular frequency is doubled (\( \omega' = 2\omega \)): \[ L' = I \cdot (2\omega) \] - The new angular momentum \( L' \) is: \[ L' = 2L \] 4. Changes in Kinetic Energy: - If the kinetic energy is halved: \[ K' = \frac{K}{2} = \frac{1}{2} I (2\omega)^2 \] - Simplifying, \[ K' = \frac{1}{2} I (4\omega^2) = 4 \cdot \frac{1}{2} I \omega^2 \] - This is equivalent to four times the original kinetic energy \( K \)., 5. New Angular Momentum Calculation: - Since the kinetic energy is quadrupled, and angular momentum is directly proportional to the angular velocity (\( \omega \)), \[ L' = \frac{L}{4} \] Conclusion:
- When the angular frequency of a rotating particle is doubled and its kinetic energy is halved, its angular momentum decreases by a factor of 4. - The new angular momentum is \( L/4 \). Thus, the correct answer is **(b)**.
1. Initial Conditions: - Let the initial angular momentum of the particle be \( L \). - The angular frequency (ω) of rotation is given by: \[ \omega = \frac{2\pi}{T} \] where \( T \) is the period of rotation. 2. Angular Momentum: - Angular momentum \( L \) is given by: \[ L = I\omega \] where \( I \) is the moment of inertia of the particle. - The kinetic energy \( K \) of the rotating particle is: \[ K = \frac{1}{2} I \omega^2 \] 3. Changes in Angular Frequency: - If the angular frequency is doubled (\( \omega' = 2\omega \)): \[ L' = I \cdot (2\omega) \] - The new angular momentum \( L' \) is: \[ L' = 2L \] 4. Changes in Kinetic Energy: - If the kinetic energy is halved: \[ K' = \frac{K}{2} = \frac{1}{2} I (2\omega)^2 \] - Simplifying, \[ K' = \frac{1}{2} I (4\omega^2) = 4 \cdot \frac{1}{2} I \omega^2 \] - This is equivalent to four times the original kinetic energy \( K \)., 5. New Angular Momentum Calculation: - Since the kinetic energy is quadrupled, and angular momentum is directly proportional to the angular velocity (\( \omega \)), \[ L' = \frac{L}{4} \] Conclusion:
- When the angular frequency of a rotating particle is doubled and its kinetic energy is halved, its angular momentum decreases by a factor of 4. - The new angular momentum is \( L/4 \). Thus, the correct answer is **(b)**.
2. A car is moving with a speed of 108 km/hr on a circular path of radius 500 m. Its speed is increasing at the rate of 2 m/s². What is the acceleration of the car?
A. \( 9.8 \, \text{m/s}^2 \)
B. \( 2.7 \, \text{m/s}^2 \)
C. \( 3.6 \, \text{m/s}^2 \)
D. \( 1.8 \, \text{m/s}^2 \)
The correct answer is B. \( 2.7 \, \text{m/s}^2 \).
Explanation:
The acceleration of the car is a combination of its tangential acceleration and centripetal acceleration. The tangential acceleration is given as \( a_t = 2 \, \text{m/s}^2 \).
The centripetal acceleration \( a_c \) is calculated using: \[ a_c = \frac{v^2}{r} \] where \( v = 108 \, \text{km/hr} = 30 \, \text{m/s} \) and \( r = 500 \, \text{m} \). Substituting these values: \[ a_c = \frac{30^2}{500} = 1.8 \, \text{m/s}^2 \] The net acceleration \( a \) is: \[ a = \sqrt{a_t^2 + a_c^2} = \sqrt{2^2 + 1.8^2} \approx 2.7 \, \text{m/s}^2 \]
Conclusion:
The total acceleration of the car is approximately \( 2.7 \, \text{m/s}^2 \).
Explanation:
The acceleration of the car is a combination of its tangential acceleration and centripetal acceleration. The tangential acceleration is given as \( a_t = 2 \, \text{m/s}^2 \).
The centripetal acceleration \( a_c \) is calculated using: \[ a_c = \frac{v^2}{r} \] where \( v = 108 \, \text{km/hr} = 30 \, \text{m/s} \) and \( r = 500 \, \text{m} \). Substituting these values: \[ a_c = \frac{30^2}{500} = 1.8 \, \text{m/s}^2 \] The net acceleration \( a \) is: \[ a = \sqrt{a_t^2 + a_c^2} = \sqrt{2^2 + 1.8^2} \approx 2.7 \, \text{m/s}^2 \]
Conclusion:
The total acceleration of the car is approximately \( 2.7 \, \text{m/s}^2 \).
3. The moment of inertia of a uniform circular disc about an axis passing through its center is \( 6 \, \text{kg} \, \text{m}^2 \). Its moment of inertia about an axis perpendicular to its plane and just touching the rim will be:
A. \( 18 \, \text{kg} \, \text{m}^2 \)
B. \( 30 \, \text{kg} \, \text{m}^2 \)
C. \( 15 \, \text{kg} \, \text{m}^2 \)
D. \( 3 \, \text{kg} \, \text{m}^2 \)
The correct answer is (a) \( 18 \, \text{kg} \, \text{m}^2 \).
Solution:
1. **Moment of Inertia About the Center**: - The moment of inertia of a uniform circular disc about an axis passing through its center is given as: \[ I_{\text{center}} = 6 \, \text{kg} \, \text{m}^2 \] 2. **Applying the Parallel Axis Theorem**: - To find the moment of inertia about an axis perpendicular to the plane and just touching the rim, we use the Parallel Axis Theorem: \[ I_{\text{rim}} = I_{\text{center}} + M R^2 \] where: - \( I_{\text{center}} \) is the moment of inertia about the center, - \( M \) is the mass of the disc, - \( R \) is the radius of the disc. 3. **Determining Mass and Radius**: - Since \( I_{\text{center}} = \frac{1}{2} M R^2 \), we can write: \[ M R^2 = 2 \cdot I_{\text{center}} \] Substituting \( I_{\text{center}} = 6 \, \text{kg} \, \text{m}^2 \): \[ M R^2 = 12 \, \text{kg} \, \text{m}^2 \] 4. **Finding \( I_{\text{rim}} \)**: - Substituting into the Parallel Axis Theorem: \[ I_{\text{rim}} = I_{\text{center}} + M R^2 \] \[ I_{\text{rim}} = 6 + 12 = 18 \, \text{kg} \, \text{m}^2 \] Conclusion:
- The moment of inertia of the disc about an axis perpendicular to its plane and just touching the rim is \( 18 \, \text{kg} \, \text{m}^2 \). - Thus, the correct answer is **(a)**.
Solution:
1. **Moment of Inertia About the Center**: - The moment of inertia of a uniform circular disc about an axis passing through its center is given as: \[ I_{\text{center}} = 6 \, \text{kg} \, \text{m}^2 \] 2. **Applying the Parallel Axis Theorem**: - To find the moment of inertia about an axis perpendicular to the plane and just touching the rim, we use the Parallel Axis Theorem: \[ I_{\text{rim}} = I_{\text{center}} + M R^2 \] where: - \( I_{\text{center}} \) is the moment of inertia about the center, - \( M \) is the mass of the disc, - \( R \) is the radius of the disc. 3. **Determining Mass and Radius**: - Since \( I_{\text{center}} = \frac{1}{2} M R^2 \), we can write: \[ M R^2 = 2 \cdot I_{\text{center}} \] Substituting \( I_{\text{center}} = 6 \, \text{kg} \, \text{m}^2 \): \[ M R^2 = 12 \, \text{kg} \, \text{m}^2 \] 4. **Finding \( I_{\text{rim}} \)**: - Substituting into the Parallel Axis Theorem: \[ I_{\text{rim}} = I_{\text{center}} + M R^2 \] \[ I_{\text{rim}} = 6 + 12 = 18 \, \text{kg} \, \text{m}^2 \] Conclusion:
- The moment of inertia of the disc about an axis perpendicular to its plane and just touching the rim is \( 18 \, \text{kg} \, \text{m}^2 \). - Thus, the correct answer is **(a)**.
4. A particle undergoes uniform circular motion. About which point on the plane of the circle will the angular momentum of the particle remain conserved?
A. Center of the circle
B. On the circumference of the circle
C. Inside the circle
D. Outside the circle
The correct answer is A. Center of the circle.
Explanation:
Angular momentum is conserved about a point if there is no net external torque acting about that point. For uniform circular motion, the forces are centripetal and act radially towards the center of the circle. Thus, there is no external torque about the center of the circle.
Conclusion:
The angular momentum of the particle remains conserved about the center of the circle.
Explanation:
Angular momentum is conserved about a point if there is no net external torque acting about that point. For uniform circular motion, the forces are centripetal and act radially towards the center of the circle. Thus, there is no external torque about the center of the circle.
Conclusion:
The angular momentum of the particle remains conserved about the center of the circle.
5. Two particles \( A \) and \( B \), initially at rest, move towards each other under a mutual force of attraction. At the instant when the speed of \( A \) is \( u \) and the speed of \( B \) is \( 2u \), the speed of the center of mass is:
A. Zero
B. \( u \)
C. \( 1.5u \)
D. \( 3u \)
The correct answer is A. Zero.
Explanation:
The center of mass of a system moves according to the net external force acting on the system. Since the forces between the two particles are internal, there is no external force acting on the system. Therefore, the velocity of the center of mass remains zero.
Conclusion:
The speed of the center of mass of the system is zero.
Explanation:
The center of mass of a system moves according to the net external force acting on the system. Since the forces between the two particles are internal, there is no external force acting on the system. Therefore, the velocity of the center of mass remains zero.
Conclusion:
The speed of the center of mass of the system is zero.
6. The moment of inertia of a body about a given axis is \( 1.2 \, \text{kg} \, \text{m}^2 \). Initially, the body is at rest. In order to produce a rotating kinetic energy of \( 1500 \, \text{joules} \), an angular acceleration of \( 25 \, \text{radian/sec}^2 \) must be applied about that axis for a duration of:
A. 4 sec
B. 2 sec
C. 8 sec
D. 10 sec
The correct answer is B. 2 sec.
Explanation:
The rotational kinetic energy is given by: \[ K = \frac{1}{2} I \omega^2 \] Substituting \( K = 1500 \, \text{joules} \) and \( I = 1.2 \, \text{kg} \, \text{m}^2 \): \[ 1500 = \frac{1}{2} \cdot 1.2 \cdot \omega^2 \implies \omega^2 = 2500 \implies \omega = 50 \, \text{radian/sec} \] Using the equation of motion for angular acceleration: \[ \omega = \alpha t \] where \( \alpha = 25 \, \text{radian/sec}^2 \), we solve for \( t \): \[ t = \frac{\omega}{\alpha} = \frac{50}{25} = 2 \, \text{sec} \]
Conclusion:
The required time to achieve the given kinetic energy is \( 2 \, \text{seconds} \).
Explanation:
The rotational kinetic energy is given by: \[ K = \frac{1}{2} I \omega^2 \] Substituting \( K = 1500 \, \text{joules} \) and \( I = 1.2 \, \text{kg} \, \text{m}^2 \): \[ 1500 = \frac{1}{2} \cdot 1.2 \cdot \omega^2 \implies \omega^2 = 2500 \implies \omega = 50 \, \text{radian/sec} \] Using the equation of motion for angular acceleration: \[ \omega = \alpha t \] where \( \alpha = 25 \, \text{radian/sec}^2 \), we solve for \( t \): \[ t = \frac{\omega}{\alpha} = \frac{50}{25} = 2 \, \text{sec} \]
Conclusion:
The required time to achieve the given kinetic energy is \( 2 \, \text{seconds} \).
7. Two discs have the same mass and rotate about the same axis. \( r_1 \) and \( r_2 \) are the radii of the two discs (\( r_1 > r_2 \)). What is the relation between \( I_1 \) and \( I_2 \)?
A. \( I_1 > I_2 \)
B. \( I_1 >> I_2 \)
C. \( I_1 < I_2 \)
D. \( I_1 = I_2 \)
The correct answer is A. \( I_1 > I_2 \).
Explanation:
The moment of inertia \( I \) of a disc about its central axis is given by: \[ I = \frac{1}{2} M r^2 \] Since the masses (\( M \)) of the two discs are the same, the moment of inertia depends on the square of the radius. As \( r_1 > r_2 \), it follows that: \[ I_1 = \frac{1}{2} M r_1^2 > I_2 = \frac{1}{2} M r_2^2 \]
Conclusion:
The disc with the larger radius has a greater moment of inertia.
Explanation:
The moment of inertia \( I \) of a disc about its central axis is given by: \[ I = \frac{1}{2} M r^2 \] Since the masses (\( M \)) of the two discs are the same, the moment of inertia depends on the square of the radius. As \( r_1 > r_2 \), it follows that: \[ I_1 = \frac{1}{2} M r_1^2 > I_2 = \frac{1}{2} M r_2^2 \]
Conclusion:
The disc with the larger radius has a greater moment of inertia.
8. The kinetic energy of a body is \( 4 \, \text{joules} \) and its moment of inertia is \( 2 \, \text{kg} \, \text{m}^2 \). What is the angular momentum of the body?
A. \( 4 \, \text{kg} \, \text{m}^2/\text{sec} \)
B. \( 5 \, \text{kg} \, \text{m}^2/\text{sec} \)
C. \( 6 \, \text{kg} \, \text{m}^2/\text{sec} \)
D. \( 7 \, \text{kg} \, \text{m}^2/\text{sec} \)
The correct answer is A. \( 4 \, \text{kg} \, \text{m}^2/\text{sec} \).
Explanation:
The rotational kinetic energy is related to the angular momentum \( L \) and moment of inertia \( I \) as: \[ K = \frac{L^2}{2I} \] Substituting \( K = 4 \, \text{joules} \) and \( I = 2 \, \text{kg} \, \text{m}^2 \): \[ 4 = \frac{L^2}{2 \cdot 2} \implies L^2 = 16 \implies L = 4 \, \text{kg} \, \text{m}^2/\text{sec} \]
Conclusion:
The angular momentum of the body is \( 4 \, \text{kg} \, \text{m}^2/\text{sec} \).
Explanation:
The rotational kinetic energy is related to the angular momentum \( L \) and moment of inertia \( I \) as: \[ K = \frac{L^2}{2I} \] Substituting \( K = 4 \, \text{joules} \) and \( I = 2 \, \text{kg} \, \text{m}^2 \): \[ 4 = \frac{L^2}{2 \cdot 2} \implies L^2 = 16 \implies L = 4 \, \text{kg} \, \text{m}^2/\text{sec} \]
Conclusion:
The angular momentum of the body is \( 4 \, \text{kg} \, \text{m}^2/\text{sec} \).
9. A mass is revolving in a circle which is in the plane of the paper. The direction of angular acceleration is:
A. Upward to the radius
B. Towards the radius
C. Tangential
D. At right angles to angular velocity
The correct answer is C. Tangential.
Explanation:
Angular acceleration occurs when there is a change in angular velocity with respect to time. For a mass moving in a circular path, the centripetal force acting towards the center of the circle causes angular acceleration. The direction of the angular acceleration is Tangential to the radius of the circle.
Conclusion:
The direction of angular acceleration is Tangential to the radius the circular path.
Explanation:
Angular acceleration occurs when there is a change in angular velocity with respect to time. For a mass moving in a circular path, the centripetal force acting towards the center of the circle causes angular acceleration. The direction of the angular acceleration is Tangential to the radius of the circle.
Conclusion:
The direction of angular acceleration is Tangential to the radius the circular path.
10. By keeping the moment of inertia of a body constant, if we double the time period, then the angular momentum of the body:
A. Remains constant
B. Becomes half
C. Doubles
D. Quadruples
The correct answer is (b) Becomes half.
Solution:
1. **Angular Momentum Relation**:
The angular momentum \( L \) of a rotating body is related to its moment of inertia \( I \) and angular velocity \( \omega \) by: \[ L = I \omega \] Here, the moment of inertia \( I \) is constant.
2. **Angular Velocity and Time Period**:
The angular velocity \( \omega \) is related to the time period \( T \) of the rotation by: \[ \omega = \frac{2\pi}{T} \] If the time period \( T \) doubles (\( T' = 2T \)), the new angular velocity \( \omega' \) becomes: \[ \omega' = \frac{2\pi}{T'} = \frac{2\pi}{2T} = \frac{\omega}{2} \]
3. **New Angular Momentum**:
Substituting \( \omega' = \frac{\omega}{2} \) into the angular momentum equation: \[ L' = I \cdot \omega' = I \cdot \frac{\omega}{2} = \frac{L}{2} \] Thus, the angular momentum \( L \) becomes half of its original value when the time period is doubled.
Conclusion:
- When the time period \( T \) of a rotating body is doubled while keeping the moment of inertia constant, the angular momentum becomes half of its original value.
- The correct answer is: \[ \boxed{\text{(b) Becomes half}} \]
Solution:
1. **Angular Momentum Relation**:
The angular momentum \( L \) of a rotating body is related to its moment of inertia \( I \) and angular velocity \( \omega \) by: \[ L = I \omega \] Here, the moment of inertia \( I \) is constant.
2. **Angular Velocity and Time Period**:
The angular velocity \( \omega \) is related to the time period \( T \) of the rotation by: \[ \omega = \frac{2\pi}{T} \] If the time period \( T \) doubles (\( T' = 2T \)), the new angular velocity \( \omega' \) becomes: \[ \omega' = \frac{2\pi}{T'} = \frac{2\pi}{2T} = \frac{\omega}{2} \]
3. **New Angular Momentum**:
Substituting \( \omega' = \frac{\omega}{2} \) into the angular momentum equation: \[ L' = I \cdot \omega' = I \cdot \frac{\omega}{2} = \frac{L}{2} \] Thus, the angular momentum \( L \) becomes half of its original value when the time period is doubled.
Conclusion:
- When the time period \( T \) of a rotating body is doubled while keeping the moment of inertia constant, the angular momentum becomes half of its original value.
- The correct answer is: \[ \boxed{\text{(b) Becomes half}} \]
11. If a horizontal cylindrical tube, partly filled with water, is rapidly rotated about a vertical axis passing through its center, the moment of inertia of the water about its axis will:
A. Decrease
B. Increase
C. Not change
D. Increase or decrease depending upon clockwise or anticlockwise sense of rotation
The correct answer is B. Increase.
Explanation:
When a rotating system accelerates, such as a cylindrical tube filled with water, the water behaves like a rotating mass. As the tube rotates rapidly, the water experiences a centrifugal force pushing it outwards. This increases the effective radius of rotation and hence the moment of inertia.
Conclusion:
The moment of inertia of the water increases as it rotates faster.
Explanation:
When a rotating system accelerates, such as a cylindrical tube filled with water, the water behaves like a rotating mass. As the tube rotates rapidly, the water experiences a centrifugal force pushing it outwards. This increases the effective radius of rotation and hence the moment of inertia.
Conclusion:
The moment of inertia of the water increases as it rotates faster.
12. The moment of inertia of a copper disc, rotating about an axis passing through its center and perpendicular to its plane:
A. Increases if its temperature is increased
B. Changes if its axis of rotation is changed
C. Increases if its angular velocity is increased
D. Both (a) and (b) are correct
The correct answer is D. Both (a) and (b) are correct.
Explanation:
(a) The moment of inertia of a body depends on the distribution of mass relative to the axis of rotation. As the temperature of a material like copper increases, the atoms vibrate more, which can alter the distribution of mass and hence affect the moment of inertia.
(b) Changing the axis of rotation can change the moment of inertia, as seen when a disc rotates around its center compared to an axis through the rim.
Conclusion:
Both an increase in temperature and a change in axis of rotation affect the moment of inertia of a copper disc.
Explanation:
(a) The moment of inertia of a body depends on the distribution of mass relative to the axis of rotation. As the temperature of a material like copper increases, the atoms vibrate more, which can alter the distribution of mass and hence affect the moment of inertia.
(b) Changing the axis of rotation can change the moment of inertia, as seen when a disc rotates around its center compared to an axis through the rim.
Conclusion:
Both an increase in temperature and a change in axis of rotation affect the moment of inertia of a copper disc.
13. The kinetic energy of a body is \( 4 \, \text{joules} \) and its moment of inertia is \( 2 \, \text{kg} \, \text{m}^2 \). What is the angular momentum of the body?
A. \( 4 \, \text{kg} \, \text{m}^2/\text{sec} \)
B. \( 5 \, \text{kg} \, \text{m}^2/\text{sec} \)
C. \( 6 \, \text{kg} \, \text{m}^2/\text{sec} \)
D. \( 7 \, \text{kg} \, \text{m}^2/\text{sec} \)
The correct answer is A. \( 4 \, \text{kg} \, \text{m}^2/\text{sec} \).
Explanation:
The angular momentum \( L \) of a rotating body is given by: \[ L = I \omega \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. First, find \( \omega \) using the kinetic energy: \[ K = \frac{1}{2} I \omega^2 \] Given \( K = 4 \, \text{joules} \) and \( I = 2 \, \text{kg} \, \text{m}^2 \): \[ 4 = \frac{1}{2} \cdot 2 \cdot \omega^2 \implies \omega^2 = 4 \implies \omega = 2 \, \text{rad/sec} \] Now, find \( L \): \[ L = I \omega = 2 \cdot 2 = 4 \, \text{kg} \, \text{m}^2/\text{sec} \]
Conclusion:
The angular momentum of the body is \( 4 \, \text{kg} \, \text{m}^2/\text{sec} \).
Explanation:
The angular momentum \( L \) of a rotating body is given by: \[ L = I \omega \] where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. First, find \( \omega \) using the kinetic energy: \[ K = \frac{1}{2} I \omega^2 \] Given \( K = 4 \, \text{joules} \) and \( I = 2 \, \text{kg} \, \text{m}^2 \): \[ 4 = \frac{1}{2} \cdot 2 \cdot \omega^2 \implies \omega^2 = 4 \implies \omega = 2 \, \text{rad/sec} \] Now, find \( L \): \[ L = I \omega = 2 \cdot 2 = 4 \, \text{kg} \, \text{m}^2/\text{sec} \]
Conclusion:
The angular momentum of the body is \( 4 \, \text{kg} \, \text{m}^2/\text{sec} \).
14. The total energy of a rolling ring of mass \( m \) and radius \( R \):
A. \( \frac{3}{2} mv^2 \)
B. \( \frac{1}{2} mv^2 \)
C. \( mv^2 \)
D. \( \frac{5}{2} mv^2 \)
The correct answer is B. \( \frac{1}{2} mv^2 \).
Explanation:
The total energy of a rolling ring consists of both translational and rotational kinetic energy: \[ E = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \] For a rolling ring, \( I = mr^2 \) and \( \omega = \frac{v}{R} \): \[ E = \frac{1}{2} m v^2 + \frac{1}{2} m \left(\frac{v}{R}\right)^2 R^2 \] Simplifying, this leads to: \[ E = m v^2 \]
Conclusion:
The total energy of the rolling ring is \( mv^2 \).
Explanation:
The total energy of a rolling ring consists of both translational and rotational kinetic energy: \[ E = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \] For a rolling ring, \( I = mr^2 \) and \( \omega = \frac{v}{R} \): \[ E = \frac{1}{2} m v^2 + \frac{1}{2} m \left(\frac{v}{R}\right)^2 R^2 \] Simplifying, this leads to: \[ E = m v^2 \]
Conclusion:
The total energy of the rolling ring is \( mv^2 \).
15. Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the center of mass is:
A. 30 m/s
B. 20 m/s
C. 10 m/s
D. 5 m/s
The correct answer is C. 10 m/s.
Explanation:
The velocity of the center of mass \( V_{cm} \) is given by: \[ V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \] Given \( m_1 = 10 \, \text{kg} \), \( v_1 = 14 \, \text{m/s} \), \( m_2 = 4 \, \text{kg} \), and \( v_2 = 0 \) (since the lighter block is initially at rest): \[ V_{cm} = \frac{10 \times 14 + 4 \times 0}{10 + 4} = \frac{140}{14} = 10 \, \text{m/s} \]
Conclusion:
The velocity of the center of mass is \( 10 \, \text{m/s} \).
Explanation:
The velocity of the center of mass \( V_{cm} \) is given by: \[ V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \] Given \( m_1 = 10 \, \text{kg} \), \( v_1 = 14 \, \text{m/s} \), \( m_2 = 4 \, \text{kg} \), and \( v_2 = 0 \) (since the lighter block is initially at rest): \[ V_{cm} = \frac{10 \times 14 + 4 \times 0}{10 + 4} = \frac{140}{14} = 10 \, \text{m/s} \]
Conclusion:
The velocity of the center of mass is \( 10 \, \text{m/s} \).
16. A child is standing with folded hands at the center of a platform rotating about its central axis. The kinetic energy of the system is \( K \). The child now stretches his arms so that the moment of inertia of the system doubles. The kinetic energy of the system now is:
A. \( K/2 \)
B. \( 2K \)
C. \( K/4 \)
D. \( 4K \)
The correct answer is A. \( K/2 \).
Explanation:
The kinetic energy of a rotating body is given by: \[ K = \frac{1}{2} I \omega^2 \] If the moment of inertia \( I \) doubles, then: \[ K' = \frac{1}{2} (2I) \omega^2 = K/2 \]
Conclusion:
The kinetic energy is halved when the moment of inertia doubles.
Explanation:
The kinetic energy of a rotating body is given by: \[ K = \frac{1}{2} I \omega^2 \] If the moment of inertia \( I \) doubles, then: \[ K' = \frac{1}{2} (2I) \omega^2 = K/2 \]
Conclusion:
The kinetic energy is halved when the moment of inertia doubles.
17. Angular momentum is:
A. An axial vector
B. A polar vector
C. A scalar as well as a vector
D. A polar vector and axial vector both
The correct answer is A. An axial vector.
Explanation:
Angular momentum is an axial vector because it is directed along the axis of rotation. It has both magnitude and direction, following the right-hand rule.
Conclusion:
Angular momentum is an axial vector.
Explanation:
Angular momentum is an axial vector because it is directed along the axis of rotation. It has both magnitude and direction, following the right-hand rule.
Conclusion:
Angular momentum is an axial vector.
18. A particle of mass \( m \) is moving with a constant velocity along a line parallel to the +ve direction of the X-axis. The magnitude of its angular momentum with respect to the origin:
A. Is zero
B. Goes on increasing as \( x \) is increased
C. Goes on decreasing as \( x \) is increased
D. Remains constant for all positions of the particle
The correct answer is A. Is zero.
Explanation:
Angular momentum is calculated as \( L = r \times p \), where \( r \) is the position vector from the origin to the particle and \( p \) is its linear momentum. Since the particle is moving with constant velocity along the X-axis, its position vector does not change direction with time (i.e., no rotation involved), thus angular momentum with respect to the origin is zero.
Conclusion:
The angular momentum of a particle moving in a straight line is zero.
Explanation:
Angular momentum is calculated as \( L = r \times p \), where \( r \) is the position vector from the origin to the particle and \( p \) is its linear momentum. Since the particle is moving with constant velocity along the X-axis, its position vector does not change direction with time (i.e., no rotation involved), thus angular momentum with respect to the origin is zero.
Conclusion:
The angular momentum of a particle moving in a straight line is zero.
19. A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass the same, which one of the following will not be affected?
A. Moment of inertia
B. Angular momentum
C. Angular velocity
D. Rotational kinetic energy
The correct answer is B. Angular momentum.
Explanation:
The moment of inertia \( I \) of a solid sphere is: \[ I = \frac{2}{5} m r^2 \] When the radius \( r \) is increased, the moment of inertia changes accordingly. However, the angular momentum \( L \) depends on the moment of inertia and angular velocity (\( \omega \)): \[ L = I \omega \] Since the mass and moment of inertia change proportionally (for a solid sphere, \( I \propto r^2 \)), but the angular velocity remains the same because the sphere is in free space, the angular momentum does not change.
Conclusion:
Changing the radius of a solid sphere affects its moment of inertia, angular velocity, and rotational kinetic energy, but not its angular momentum.
Explanation:
The moment of inertia \( I \) of a solid sphere is: \[ I = \frac{2}{5} m r^2 \] When the radius \( r \) is increased, the moment of inertia changes accordingly. However, the angular momentum \( L \) depends on the moment of inertia and angular velocity (\( \omega \)): \[ L = I \omega \] Since the mass and moment of inertia change proportionally (for a solid sphere, \( I \propto r^2 \)), but the angular velocity remains the same because the sphere is in free space, the angular momentum does not change.
Conclusion:
Changing the radius of a solid sphere affects its moment of inertia, angular velocity, and rotational kinetic energy, but not its angular momentum.
20. Two circular discs A and B have equal masses and uniform thickness but have densities \( \rho_1 \) and \( \rho_2 \) such that \( \rho_1 > \rho_2 \). Their moment of inertia is:
A. \( I_1 < I_2 \)
B. \( I_1 > I_2 \)
C. \( I_1 \leq I_2 \)
D. \( I_1 = I_2 \)
The correct answer is A. \( I_1 < I_2 \).
Explanation:
The moment of inertia of a disc is given by: \[ I = \frac{1}{2} M R^2 \] where \( M \) is the mass and \( R \) is the radius. If two discs have equal mass but different densities: \[ \text{Disc A:} \quad M_1 = \rho_1 V_1, \quad R_1 \] \[ \text{Disc B:} \quad M_2 = \rho_2 V_2, \quad R_2 \] Since \( \rho_1 > \rho_2 \) and \( V \propto R^2 \), it follows: \[ I_1 = \frac{1}{2} M_1 R_1^2 < \frac{1}{2} M_2 R_2^2 = I_2 \]
Conclusion:
The disc with higher density will have a smaller moment of inertia compared to the one with lower density for equal masses.
Explanation:
The moment of inertia of a disc is given by: \[ I = \frac{1}{2} M R^2 \] where \( M \) is the mass and \( R \) is the radius. If two discs have equal mass but different densities: \[ \text{Disc A:} \quad M_1 = \rho_1 V_1, \quad R_1 \] \[ \text{Disc B:} \quad M_2 = \rho_2 V_2, \quad R_2 \] Since \( \rho_1 > \rho_2 \) and \( V \propto R^2 \), it follows: \[ I_1 = \frac{1}{2} M_1 R_1^2 < \frac{1}{2} M_2 R_2^2 = I_2 \]
Conclusion:
The disc with higher density will have a smaller moment of inertia compared to the one with lower density for equal masses.
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