A particle moves 3.0 m along a circle of radius 1.5 m. (a) Through what angle does it rotate? (b) If the particle makes this trip in 1.0 s at a constant speed, what is its angular velocity? (c) What is its acceleration?

(a) 114.65° or 2.00 rad

(b) 2.00 rad/s

(c) 0 m/s²

Explanation:

(a)

Note: As the particle moves along the circle, it forms an arc.

Length of an arc = (Ф/360)×2πr

L = (Ф/360)×2πr Equation 1

Where L = distance at which the particle moves along the circle, Ф = angle of rotation of the particle, r = radius of the circle.

make Ф the subject of the equation

Ф = 360L/2πr Equation 2

Given: L = 3 m, r = 1.5 m

Substitute into equation 2

Ф = 360(3)/(2×3.14×1.5)

Ф = 1080/9.42

Ф = 114.65° or 2.00 rad

(b)

Angular velocity(ω) = Ф/t

ω = ΔФ/Δt Equation 3

Where ω = angular velocity, Δt = time taken to make the trip, ΔФ = angle through which the particle rotates

Given: ΔФ = 2.00 rad, Δt = 1.0 s

Substitute into equation 3

ω = 2.00/1

ω = 2.00 rad/s

(c)

α = dω/dt Equation 4

Where α = angular acceleration, dω = change in velocity, dt = change in time

α = d(2.00)/dt

Note: since ω is a constant, and the differentiation of a constant is equal to zero,

therefore,

α = 0 rad/s².

But,

a = αr Equation 5

Where a = acceleration of the particle

Given: α = rad/s², r = 1.5 m

Substitute into equation 5

a = 0(1.5)

a = 0 m/s²

answer: decibel

explanation:

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000