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mikau

Member

Registered: 2005-08-22

Posts: 1,504

newtons law for torque and angular momentum, really could use help

curses, I'm confounded.

okay,  currently learning about angular momentum, which, in the case of a single particle, is given as

L = r x p, where r is the position vector of the particle, and p =  momentum of the particle = mv, where m is the mass of the particle and v is  the velocity vector of the particle.

Good so far.

but my book soon defines the angular momentum of a system of particles to be:

L = ∑ L[sub]i[/sub] where L[sub]i[/sub] is the angular momentum of the i'th particle.

still okay.

now they show that dL/dt  = net Torque of the system

L = ∑ L[sub]i[/sub] = ∑ m (r[sub]i[/sub]) x (v[sub]i[/sub])

differentiating with respect to t yields:

dL/dt =  ∑ m(r[sub]i[/sub])x(dv[sub]i[/sub] /dt) + m(v[sub]i[/sub])x(dr[sub]i[/sub]/dt)

now we know dr[sub]i[/sub]/dt is the change in the position vector over dt, which is the same as the velocity vector v[sub]i[/sub], substituting this gives us:

dL/dt =  ∑ m(r[sub]i[/sub])x(dv[sub]i[/sub] /dt) + m(v[sub]i[/sub])x(v[sub]i[/sub])

by definition of cross product, (v[sub]i[/sub])x(v[sub]i[/sub]) is zero, furthermore, we can replace dv[sub]i[/sub]/dt with a[sub]i[/sub], the acceleration of the i'th particle, so we obtain:

dL/dt =  ∑ m(r[sub]i[/sub])x(a[sub]i[/sub])   + 0

which we can also write as

dL/dt = ∑ (r[sub]i[/sub])x(m*a[sub]i[/sub])  and sustituting F[sub]net,i[/sub] for m*a[sub]i[/sub], we obtain

dL/dt = ∑ (r[sub]i[/sub])x(F[sub]net,i[/sub])  but this is exactly how we define the net torque on the system about the axis of rotation O where O is at the tale of r, the position vector.

And so we get dL/dt = net torque, or the rate of change in the angular momentum of the system about O equals the net torque of the system about O.

That all makes perfect sense, until my book gave this word of caution, which i quote word for word here (comments by me are made in brackets []:

this equation [net torque = dL/dt] is analogous to Fnet = dP/dt [newtons law for linear momentum] but requires an extra caution: torques and the system's angular momentum must be measured relative to the same origin.[makes sense!] If the center of mass of the system is not accelerating relative to an inertial frame, that origin can be any point. [...um] However, if the center of mass of the system is accelerating, the origin can be only at the center of mass. [WHAT? WHY?] As an example, consider a wheel as the system of particles. If the wheel is rotating about an axis, that is fixed relative to the ground, then the origin for applying the equation  [net torque = dL/dt]  can be any point that is stationary relative to the ground. However, if the wheel is rotating about an axis that is accelerating (such as when the wheel rolls down a ramp) then the origin can only be at the wheels center of mass.

and they've lost me with that. I see no reason why we can't, for instance, apply the same reasoning to find the net torque about a stationary origin, of a system of particles that falls freely. I can't see how an acceleration of the systems center of mass, crashes the logic behind the formula dL/dt = net torque.

a thousand pounds of chocolate to whoever can explain this to me.

Last edited by mikau (2008-03-31 16:49:09)

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