Point Particle Motion problem

[jackiemoon]

Hi,

Can anybody help with the following:

A point particle moves in a plane with trajectory given by r(t) = R [θ (t)]^3/2, where R is a constant.

The angle θ in radians increases in time according to the equation θ (t) = 1/2 α t^2,
where α is a constant whose numerical value is α = 1 s^-2.

a. Sketch the trajectory of the particle.
b. Compute the time t > 0, if any, for which the radial and tangential components of the velocity of the particle coincide.
c. Compute the time t > 0, if any, for which the acceleration of the particle is purely tangential.
d. Compute the time t > 0, if any, for which the radial acceleration of the particle is twice its tangential acceleration.

I really am struggling with this question. I'm not even sure about sketching the trajectory, which is usually the least I can do in these type of questions! Can anybody help and explain it in really simple terms. Thanks for any help with this.

 

[LowlyPion]

What you have is an equation in polar coordinates don't you?

And if |R| is constant, it is described how?

 

[thomate1]

Sure, I'd be happy to help with this problem. Let's break it down step by step:

a. Sketching the trajectory: To sketch the trajectory of the particle, we need to plot the position of the particle at different times. We can do this by plugging in different values for t into the given equation for θ (t), and then using those values to calculate the corresponding values for r(t). Since the angle θ increases with time, the trajectory of the particle will spiral outward from the origin. The shape of the spiral will depend on the value of R, but it will always be a spiral.

b. Finding the time t for which the radial and tangential components of velocity coincide: To find this time, we need to set the radial and tangential components of velocity equal to each other. The radial component of velocity is given by dr/dt = R [3/2 * θ'(t) * θ(t)^1/2], and the tangential component is given by rdθ/dt = R [3/2 * θ(t)^1/2 * θ'(t)]. Setting these equal to each other and solving for t, we get t = ±√(2/α). Since t cannot be negative, the only valid solution is t = √(2/α). This means that at t = √(2/α), the particle's velocity will have equal radial and tangential components.

c. Finding the time t for which the acceleration is purely tangential: To find this time, we need to set the radial component of acceleration equal to 0. The radial component of acceleration is given by d^2r/dt^2 = R [3/2 * θ''(t) * θ(t)^1/2 + 3/4 * θ'(t)^2 * θ(t)^-1/2]. Setting this equal to 0 and solving for t, we get t = √(2/α). This means that at t = √(2/α), the particle's acceleration will be purely tangential.

d. Finding the time t for which the radial acceleration is twice the tangential acceleration: To find this time, we need to set the radial component of acceleration equal to twice the tangential component of acceleration. The radial component of acceleration is given by d^2r/dt^2 = R [