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.

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.

c) up is the direction the magnetic force is acting on the charge.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.