**Torque**

It is easier to open a door by pushing on the edge farthest from the hinges than by pushing in the middle. It is intuitive that the magnitude of the force applied and the distance from the point of application to the hinge affect the tendency of the door to rotate. This physical quantity,

**torque,**is t = r × F sin θ, where*F*is the force applied,*r*is the distance from the point of application to the center of the rotation, and θ is the angle from*r*to*F*.

**Angular Momentum****Angular momentum**is rotational momentum that is conserved in the same way that linear momentum is conserved.

For a rigid body, the angular momentum

*(L)*is the product of the moment of inertia and the angular velocity:*L*=*I*ω.For a point of mass, angular momentum can be expressed as the product of linear momentum and the radius (

*r*):*L*=*mvr*.*L*is measured in units of kilograms-meters^{2}per second or more commonly joule-seconds.The

**law of conservation of angular momentum**can be stated that the angular momentum of a system of objects is conserved if there is not external net torque acting on the system.Analogous to Newton's law (F = Δ(

*mv*)/Δ*t*) there is a rotational counterpart for rotational motion:*t*= Δ*L*/Δ*t*, or torque is the rate of change of angular momentum.Consider the example of a child who runs tangential to the edge of a playground merry-go-round with a velocity

*v*_{o}and jumps on while the merry-go-round is at rest.The only external forces are that of gravity and the contact forces provided by the support bearings, neither of which causes a torque because they are not applied to cause a horizontal rotation.

Treat the child's mass as a point of mass and the merry-go-round as a disc with a radius

*R*and mass*M*.From the conservation law, the total angular momentum of the child before the interaction is equal to the total angular momentum of the child and merry-go-round after the collision:

*mrv*_{o}=*mrv*′ +*I*ω, where*r*is the radial distance from the center of the merry-go-round to the place where the child hits. If the child jumps on the edge,*(r*=*R)*and the angular velocity for the child after the collision can be substituted for the linear velocity,*mRv*_{o}=*mR*(*R*ω)+(1/2)*MR*^{2}. If the values for the masses and the initial velocity of the child are given, the final velocity of the child and merry-go-round can be calculated.A single object may have a change in angular velocity due to the conservation of angular momentum if the distribution of the mass of the rigid body is altered.

For example, when a figure skater pulls in her extended arms, her moment of inertia will decrease, causing an increase in angular velocity. According to the conservation of angular momentum,

*I*_{ o }(ω_{ o }) =*I*_{ f }(ω_{ f }) where*I*_{ o }is the moment of inertia of the skater with arms extended,*I*_{ f }is her moment of inertia with her arms close to her body, ω_{o}is her original angular velocity, and ω_{ f }is her final angular velocity.
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