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Angular motion with constant acceleration

  • Thread starter jaded18
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  • #1
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Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time according to theta(t)=theta_0+omega_0t+alpha t^2. At time t=t_1, particle B, which also undergoes constant angular accelaration, has twice the angular acceleration, half the angular velocity, and the same angular position that particle A had at time t=0.
How long after the time t_1 does the angular velocity of B have to be to equal A's?
___
ok, here we go:
i know that i have to write expressions for the angular velocity of A and B as functions of time, and solve for t-t1???
For A, omega(t) = omega_0 + alpha(t) and
for B, omega(t) = 0.5 omega_0 + 2alpha(t-t_1)
so when i equate them and solve for t-t_1, i get something that is not the right answer!

please help :(
 

Answers and Replies

  • #2
Astronuc
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One wants expressions for [itex]\theta_A(t)[/itex] and [itex]\theta_B(t)[/itex], which are set equal at time t, so one can solve for time t, then find t - t1.

at t = 0, [itex]\theta_A(0)[/itex] = [itex]\theta_0[/itex], and at t=t1, [itex]\theta_B(t_1)[/itex] = [itex]\theta_0[/itex], which is the starting position of A at t=0.

In the expression for B, one has to address the time lag t1.
 
  • #3
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can anyone please just lay it out for me .. because i can't see it
 
Last edited:
  • #4
Astronuc
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One has for A, [tex]\theta(t)[/tex] = [tex]\theta_0\,+\,\omega_0t+\alpha t^2/2[/tex],

and for B, one must use t-t1 since it starts at t1, with

angular acceleration [itex]2\alpha[/itex] and angular velocity [itex]0.5\omega[/itex].

At the same position [tex]\theta_A(t)[/tex]=[tex]\theta_B(t)[/tex]
 
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  • #5
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how do i get to the answer. i know that the answer is (omega_0 + 2alpha(t_1)) / (2alpha) but i want to know how they got that. and i need to know what i did wrong in #1 box because it looks okay to me
 
  • #6
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i KNOW theta A equation and theta B equation.
 
  • #7
Astronuc
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Oops, sorry, I was solving for the same position.

If theta(t)=theta_0+omega_0t+ 0.5alpha t^2 for A, then

omega_A(t) = omega_0 + alpha *t

for B

omega_B(t) = 0.5*omega_0 + 2 alpha *(t-t1)

when one differentiates t2, the derivative is 2t.
 
Last edited:
  • #8
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you are wrong. i am sorry. i will change what i said in #1

Consider two particles A and B. The angular position of particle A, with constant angular acceleration, depends on time according to theta(t)=theta_0+omega_0t+0.5alpha t^2. At time t=t_1, particle B, which also undergoes constant angular accelaration, has twice the angular acceleration, half the angular velocity, and the same angular position that particle A had at time t=0.
How long after the time t_1 does the angular velocity of B have to be to equal A's?
___
ok, here we go:
i know that i have to write expressions for the angular velocity of A and B as functions of time, and solve for t-t1???
For A, omega(t) = omega_0 + alpha(t) and
for B, omega(t) = 0.5 omega_0 + 2alpha(t-t_1)
so when i equate them and solve for t-t_1, i get something that is not the right answer!

please help :(

IT IS 0.5 ALPHA T^2
 
  • #9
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THEREFORE For A, omega(t) = omega_0 + alpha(t) and
for B, omega(t) = 0.5 omega_0 + 2alpha(t-t_1)

now what do i do with these equations?
 
  • #10
Astronuc
Staff Emeritus
Science Advisor
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I suspect the problem is simply algebraic.

Equate the two expressions for angular velocity.

[tex]\omega_A(t) = \omega_B(t)[/tex]

[tex]\omega_0\,+\,\alpha{t}\,=\,\omega_0/2\,+\,2\alpha{(t-t_1)}[/tex]

which leads to

[tex]\omega_0/2 \,=\, 2\alpha{(t-t_1)}\,-\,\alpha{t}[/tex]

or


[tex]\omega_0/2 \,= \,\alpha{t} - 2 \alpha{t_1}[/tex]

take an alpha t_1 to the other side

[tex]\omega_0/2 + \alpha{t_1} \,= \,\alpha{t} - \alpha{t_1}[/tex]

and see where that leads one
 
Last edited:

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