# Angular motion with constant acceleration

1. Oct 14, 2007

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!

2. Oct 14, 2007

### Staff: Mentor

One wants expressions for $\theta_A(t)$ and $\theta_B(t)$, which are set equal at time t, so one can solve for time t, then find t - t1.

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

In the expression for B, one has to address the time lag t1.

3. Oct 15, 2007

can anyone please just lay it out for me .. because i can't see it

Last edited: Oct 15, 2007
4. Oct 15, 2007

### Staff: Mentor

One has for A, $$\theta(t)$$ = $$\theta_0\,+\,\omega_0t+\alpha t^2/2$$,

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

angular acceleration $2\alpha$ and angular velocity $0.5\omega$.

At the same position $$\theta_A(t)$$=$$\theta_B(t)$$

Last edited: Oct 15, 2007
5. Oct 15, 2007

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. Oct 15, 2007

i KNOW theta A equation and theta B equation.

7. Oct 15, 2007

### Staff: Mentor

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: Oct 15, 2007
8. Oct 15, 2007

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!

IT IS 0.5 ALPHA T^2

9. Oct 15, 2007

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. Oct 15, 2007

### Staff: Mentor

I suspect the problem is simply algebraic.

Equate the two expressions for angular velocity.

$$\omega_A(t) = \omega_B(t)$$

$$\omega_0\,+\,\alpha{t}\,=\,\omega_0/2\,+\,2\alpha{(t-t_1)}$$

$$\omega_0/2 \,=\, 2\alpha{(t-t_1)}\,-\,\alpha{t}$$

or

$$\omega_0/2 \,= \,\alpha{t} - 2 \alpha{t_1}$$

take an alpha t_1 to the other side

$$\omega_0/2 + \alpha{t_1} \,= \,\alpha{t} - \alpha{t_1}$$

and see where that leads one

Last edited: Oct 15, 2007