Answer:
(a) ωf= 42 rad/s
(b) θ = 220 rad
(c) at = 4 m/s² , v = 42 m/s
Explanation:
The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.
There is tangential acceleration (at ) and is constant.
We apply the equations of circular motion uniformly accelerated :
ωf= ω₀ + α*t Formula (1)
θ= ω₀*t + (1/2)*α*t² Formula (2)
at = α*R Formula (3)
v= ω*R Formula (4)
Where:
θ : angle that the body has rotated in a given time interval (rad)
α : angular acceleration (rad/s²)
t : time interval (s)
ω₀ : initial angular velocity ( rad/s)
ωf: final angular velocity ( rad/s)
R : radius of the circular path (cm)
at : tangential acceleration (m/s²)
v : tangential speed (m/s)
Data
α = 4.0 rad/s² : wheel’s angular acceleration
t = 10 s
ω₀ = 2.0 rad/s : wheel’s initial angular velocity
R = 1.0 m : wheel’s radium
(a) Wheel’s angular velocity after 10 s
We replace data in the formula (1):
ωf= ω₀ + α*t
ωf= 2 + (4)*(10)
ωf= 42 rad/s
(b) Angle that rotates the wheel in the 10 s interval
We replace data in the formula (2):
θ= ω₀*t + (1/2)*α*t²
θ= (2)*(10) + (1/2)*(4)*(10)²
θ= 220 rad
θ= 220 rad
(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval
We replace data in the Formula (3)
at = α*R = (4)(1)
at = 4 m/s²
We replace data in the Formula (4)
v= ω*R = (42)*(1)
v = 42 m/s
