
When mass is rotating in a plane about a fixed point, its angular momentum is directed along?
A) The axis of rotation
B) Line at an angle 45$^0$ to the axis of rotation
C) The radius
D) The tangent to the orbit
Answer
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Hint: The angular momentum is defined as the product of its moment of inertia and its angular velocity. It is analogous to linear momentum. In formula it is written as $\mathop L\limits^ \to $ =$\mathop r\limits^ \to \times \mathop p\limits^ \to $ where $\mathop L\limits^ \to $ denotes angular momentum, $\mathop r\limits^ \to $ is the point of rotation and always remain at inward in angular momentum and perpendicular to $\mathop v\limits^ \to $ and $\mathop p\limits^ \to $ is the momentum.
Complete step-by-step answer:
Now, the angular momentum is denoted by $\mathop L\limits^ \to $ and formula for it is $\mathop L\limits^ \to $ =$\mathop r\limits^ \to \times \mathop p\limits^ \to $
where, , $\mathop r\limits^ \to $ is the point of rotation and always remain at inward in angular momentum and perpendicular to $\mathop v\limits^ \to $, and $\mathop p\limits^ \to $ is the linear momentum of the particle and $\mathop p\limits^ \to $ =$m \times \mathop v\limits^ \to $ where m is the mass and v is the velocity
Angular momentum ($\mathop L\limits^ \to $)= ($\mathop r\limits^ \to $)×($m\mathop v\limits^ \to $ )
Cross product gives $\sin \theta $ so, ($\mathop L\limits^ \to $)= ($\mathop r\limits^ \to $) ($m\mathop v\limits^ \to $ ) $\sin \theta $
As $\mathop r\limits^ \to $is directed toward the point of rotation inward and $\mathop v\limits^ \to $ acts tangentially (stated above) thus their cross product is along the direction perpendicular to both.
That will be in the direction of the axis of rotation.
So, option A. is correct.
Additional Information:
Applications of angular momentum:
- Flywheels used in vehicles; a fascinating and practical application.
- Spin-orbit coupling of electrons in atoms; demonstrating hyper-fine spectroscopy phenomena.
- Figure-skaters pulling their arms in; they spin faster but over-all angular momentum is conserved.
- Space-craft launching with Earth-spin - gain speed but Earth fractionally loses momentum.
- Baseball batters transfer the angular momentum from their bats to the baseball - in linear form - but total momentum is conserved.
Note: Students usually do mistakes in understanding linear and angular momentum. Linear momentum is inertia of translational motion whereas angular momentum is inertia of rotational motion.
Momentum is denoted by p stated above and it is the product of mass and velocity of the object . In mathematical terms it is written as $\mathop p\limits^ \to $ =$m \times \mathop v\limits^ \to $ and it is always conserved so angular momentum will also be conserved.
Its S.I unit is kg${m^2}$ ${s^{ - 1}}$.
Complete step-by-step answer:
Now, the angular momentum is denoted by $\mathop L\limits^ \to $ and formula for it is $\mathop L\limits^ \to $ =$\mathop r\limits^ \to \times \mathop p\limits^ \to $
where, , $\mathop r\limits^ \to $ is the point of rotation and always remain at inward in angular momentum and perpendicular to $\mathop v\limits^ \to $, and $\mathop p\limits^ \to $ is the linear momentum of the particle and $\mathop p\limits^ \to $ =$m \times \mathop v\limits^ \to $ where m is the mass and v is the velocity
Angular momentum ($\mathop L\limits^ \to $)= ($\mathop r\limits^ \to $)×($m\mathop v\limits^ \to $ )
Cross product gives $\sin \theta $ so, ($\mathop L\limits^ \to $)= ($\mathop r\limits^ \to $) ($m\mathop v\limits^ \to $ ) $\sin \theta $
As $\mathop r\limits^ \to $is directed toward the point of rotation inward and $\mathop v\limits^ \to $ acts tangentially (stated above) thus their cross product is along the direction perpendicular to both.
That will be in the direction of the axis of rotation.
So, option A. is correct.
Additional Information:
Applications of angular momentum:
- Flywheels used in vehicles; a fascinating and practical application.
- Spin-orbit coupling of electrons in atoms; demonstrating hyper-fine spectroscopy phenomena.
- Figure-skaters pulling their arms in; they spin faster but over-all angular momentum is conserved.
- Space-craft launching with Earth-spin - gain speed but Earth fractionally loses momentum.
- Baseball batters transfer the angular momentum from their bats to the baseball - in linear form - but total momentum is conserved.
Note: Students usually do mistakes in understanding linear and angular momentum. Linear momentum is inertia of translational motion whereas angular momentum is inertia of rotational motion.
Momentum is denoted by p stated above and it is the product of mass and velocity of the object . In mathematical terms it is written as $\mathop p\limits^ \to $ =$m \times \mathop v\limits^ \to $ and it is always conserved so angular momentum will also be conserved.
Its S.I unit is kg${m^2}$ ${s^{ - 1}}$.
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