Respuesta :
Answer:
(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
(a)The angle does it rotate will be 114.65° or 2.00 rad.
(b)The angular velocity will be 2.00 rad/s
What is acceleration?
Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².It is a vector quantity. It requires both magnitudes as well as direction to define.
The given data in the problem is;
L is the length of arc = 3 m
r is the radius = 1.5 m
(a)The angle does it rotate will be 114.65° or 2.00 rad.
The length of the arc is givenh by;
[tex]\rm L= \frac{\phi}{360} \times 2 \pi r \\\\ \rm \phi= \frac{360 L}{2 \pi r} \\\\ \rm \phi= \frac{360 \times 3}{2 \times 3.14 \times 1.5} \\\\ \phi= \frac{1080}{9} \\\\ \rm \phi= 114.65^0[/tex]
Hence the angle does it rotate will be 114.65° or 2.00 rad.
(b)The angular velocity will be 2.00 rad/s.
The angular velocity is found by;
[tex]\rm \omega = \frac{\triangle \phi }{\triangle t} \\\\ \omega = \frac{2.00}{1} \\\\ \rm \omega = 2.00\ rad/sec[/tex]
Hence the angular velocity will be 2.00 rad/s
(c) The velue of the acceleration will be 0 m/s²
The angular acceleration is given as;
[tex]\rm a= \frac{d \omega }{dt}[/tex]
The angular velocity is constant.So the angular acceleration is 0 m/sec².
Hence the velue of the acceleration will be 0 m/s²
To learn more about acceleration refer to the link;
https://brainly.com/question/969842
