A wheel 2.00 m in diameter lies in a vertical plane androtates about its central axis with a constant angularacceleration of 4.00 rad/s2. The wheel starts at rest att =0, and the radius vector of a certain point P on therim makes an angle of 57.38 with the horizontal at thistime. At t 5 2.00 s, find (a) the angular speed of thewheel and, for point P, (b) thetangential speed, (c) thetotal acceleration, and (d) the angular position.
Question
Answer: a) 20.8 rad/s, b) 20.8 m/s, c) 432.67 m/s²,
d)54.08 rad.
Explanation: diameter of wheel is 2m hence radius is 2/2 = 1m.
Constant angular acceleration = α = 4.00 rad/s²
Since the motion of the wheel is of a constant angular acceleration, hence newton’s laws of motion is applicable.
Note that the body is starting from rest hence, initial angular velocity (ω0) is zero.
Time taken = 5.2s
a)
Recall that
ω = ω0 + αt
ω = 0 + 4(5.2)
ω = 20.8 rad/s.
b)
Tangential speed (v) is the linear speed and is given as
v = ωr
Where r is the radius of the wheel.
Note that ω = 20.8 rad/s
v = 20.8 × 1
v = 20.8 m/s.
c)
Total acceleration = √(a*)² + (a’)²
Where a* is the radial component of acceleration = v²/r and a’ is the vertical component of acceleration = αr.
Radial component of acceleration = v²/r = 20.8²/1 = 432.64 rad/s²
Vertical component of acceleration = αr = 4 × 1 = 4m/s²
Total acceleration = √(432.64)² + (4)²
Total acceleration = √187,177.3696 + 16
Total acceleration = √187,193.3696
Total acceleration = 432.67 m/s²
d)
θ = ω0t + αt²/2
But ω0 = 0 and θ = angular displacement ( angular position)
θ = 4(5.2)²/2
θ = 54.08 rad.