Question

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.

Answer 1

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.