Indefinite Integral for Torque

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  • #1

Homework Statement



A time-dependent torque given by a + bsin(ct) is applied to an object that's initially stationary but is free to rotate. Here a, b, and c are constants. Find an expression for the object's angular momentum as a function of time, assuming the torque is first applied at t = 0.

Homework Equations



∫a +bsin(ct) (from 0 to infinity )
τ = dL / dT

The Attempt at a Solution



Not really sure where to start, since I really don't have much experience with these types of problems. I think that this is an infinite integral problem, at least, since it only defines one point in time. Can anyone point me in the right direction? Thanks!
 

Answers and Replies

  • #2
Spinnor
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This is late and may be of no use. Suppose the problem was changed a little, instead of a torque let there be a force that acts on a mass as a function of time given by the same formula,

F = a + bsin(ct)

then F = ma = a + bsin(ct)

Could you integrate that?
 
  • #3
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0

Homework Statement



A time-dependent torque given by a + bsin(ct) is applied to an object that's initially stationary but is free to rotate. Here a, b, and c are constants. Find an expression for the object's angular momentum as a function of time, assuming the torque is first applied at t = 0.

Homework Equations



∫a +bsin(ct) (from 0 to infinity )
τ = dL / dT

The Attempt at a Solution



Not really sure where to start, since I really don't have much experience with these types of problems. I think that this is an infinite integral problem, at least, since it only defines one point in time. Can anyone point me in the right direction? Thanks!
You set up the integral correctly but just integrate from 0 up until whatever time you're up to now, t.

∫a +bsin(ct) sum of the integral is sum of the parts right?

∫a +∫bsin(ct)

Can you integrate that?
 
  • #4
386
5
I think it's not called indefinite integral but a improper integral. Improper integra is a integral that has lower limit of a specific number and a upper limit of a variable. For example, for the work done by a constant force at displacement s is W=∫F·ds=F·s (with limit from 0 to s)=F·s-F·0=Fs-0=Fs.

Apply the same principle to this equation. Since integral of torque ∫τ dt=∫r×F dt=r×Δp, which is the change in angular momentum ΔJ.

Angular momentum J is initial J(0) plus change ΔJ which is J=J(0)+ΔJ=J(0)+∫τ dt (from 0 to t). Since initially it's at rest, thus J=∫τ dt (from 0 to t).
 
  • #5
390
1
aftershock got it, you don't want to integrate up to infinity unless you wanted the angular momentum at t=infinity. Your answer for angular momentum at some time t is...
[tex]
L = \int_0^t a + b \sin (cq)\ dq
[/tex]
 

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