Question

Two particles start from point Pzero of a circle with radius R with the same initial velocity v(0) at the same time. Their E1 and E2 angular accelerations are constant.

Data:
v(0) = 5 m/s
R= 2 m
E1 = 2 rad/s(square)
E2 = −2 rad/s(square)
a, Calculate the elapsed time (t*) between the start of the particles and the time when
particle 1 catches up with particle 2. Calculate the covered distances and angular
displacements of the particles during the above time.
b, Calculate the peripheral and angular velocities of the particles at the t* moment.
c, Calculate the magnitude of their tangential and normal accelerations at the same
moment.

Answer 1

To calculate the moment when two particles with opposite angular accelerations catch up to each other, we use kinematic equations for angular motion. The peripheral and angular velocities at that moment are derived using the initial conditions and the given angular accelerations. The tangential and normal accelerations are calculated based on the particles' velocity and the radius of the circle.

Step-by-step explanation:

The calculation required to determine the time t* at which particle 1 catches up with particle 2 involves kinematic equations for angular motion. Since the initial angular velocities are equal and the angular accelerations are equal in magnitude but opposite in sign, particle 1 will catch up with particle 2 when they have rotated through the same angular displacement. The kinematic equation for angular displacement θ with constant angular acceleration is θ = ω0t + 0.5αt2, where ω0 is the initial angular velocity and α is the angular acceleration.

To find the peripheral and angular velocities at moment t*, we use ω = ω0 + αt and v = rω. The tangential and normal accelerations are given by at = rα and an = ω2r, respectively. In the student's question, tangential acceleration would vary due to changing angular velocity, while normal acceleration (centripetal acceleration) depends on the speed at which the particle is moving around the circle.