# Motion of a particle

The motion of a particle moving in a circle in the x-y plane is described by the equations: r(t)=8.27, Θ(t)=8.58t
Where Θ is the polar angle measured counter-clockwise from the + x-axis in radians, and r is the distance from the origin in m.
a)Calculate the y-coordinate of the particle at the time 1.60 s.

b)Calculate the x-component of the velocity at the time 1.90 s?

c)Calculate the magnitude of the acceleration of the particle at the time 3.70 s?

d)Calculate the x-component of the acceleration at the time 3.80s?

My teacher gave us a key to solve these but i can't make sense of it.

Part A
y = r(t)*sin(Θ(t)*t)

Part B:
vx = Θ*r*cos(Θ(t))

Part C:
ax= -(Θ^2)*r cos (Θ(t))
ay = -(Θ^2)*r sin (Θ(t))
a = sqrt(ax^2 + ay^2)

Part D:
ay = -(Θ^2)*r*sin(Θ(t))

I'm not sure what the difference is between Θ and Θ(t) & r and r(t)

## Answers and Replies

1) At time t=0 , Θ(t)=0 , with time the r(t) remains the same becaus eit is indpendent of time. So the oarticle starts with Θ(t)=0 , and with time Θ(t) increases linearly with 't'.
So at some time t , the particle moves through an angle Θ(t)=8.58t.Put the value of t=1.6/1.9 seconds , therefore now the x\y-coordinate of the particle's position is the component of r(t) over x and y axis respecticely.

You need to double differentiate r(t) for acceleration on x axis and y axis seperately and then calculate the resultant from these.

difference between Θ and Θ(t) & r and r(t)

Θ --- Symbol For Angle
Θ(t)--- Symbol For Time Dependent Angle (which changes with time)
r----- Symbol of arm length/radius of the particle's circle
r(t)----- Symbol for time-dependent radius , but here as you can see that r(t) has an expression independent of time , so it wont change with time.

BJ

or you can take and use vectors. you have r(t) = 8.27 and ~(t) = 8.58t

take 8.58 * 1.6 and you get 13.728 rad

then you have a 2d vector [8.27, 13.728]

now just convert them to rectangular coords.

x = r cos @ y = r sin @