Question

2. A rigid disk with radius 2 m is spinning at an angular speed of 8 1/s (or 8 rad/s). A constant angular acceleration is applied that slows downthe disk at a rate of 2 1/s²(or 2 rad/s²).

(i) Calculate the number of revolutions the disk does before coming to a full stop.

(ii) Calculate the centripetal acceleration of a point, positioned at 1 m from the disk rotation axis, at t = 2 s.

(iii) Calculate the tangent speed of a point positioned at the edge of the disk at t = 2 s.

Answer 1

Initial angular speed = ₀ = 8 rad/s

Final angular speed = _f = 0 rad/s

Angular acceleration = = -2 rad/s² (Negative as it is slowing the disk that is decelarating it)

Radius of the disk = R = 2m

Time taken to stop the disk = t₀

_f = ₀ + t₀

0 = 8 + (-2)t₀

t₀ = 4 sec

Angle rotated by the disk before coming to a stop =

= ₀t₀ + (t₀²)/2

= (8)(4) + (-2)(4²)/2

= 16 radians

Number of revolutions made by the disk before coming to a stop = N

N = /2

N = 16/2

N = 2.546 revolutions

Angular velocity at t=2sec = ₂

₂ = ₀ + t

₂ = 8 + (-2)(2)

₂ = 4 rad/s

Centripetal acceleration of a point positioned at 1m from the disk rotation axis at t=2 sec = a

r = 1 m

a = (₂)²r

a = (4)²(1)

a = 16 m/s²

Tangential speed of a point positioned at the edge of the disk at t=2 sec = V

V = ₂R

V = (4)(2)

V = 8 m/s

i) Number of revolutions the disk does before coming to a stop = 2.546

ii) Centripetal acceleration of a point positioned at 1m from the disk rotation axis at t=2s = 16 m/s²

iii) Tangential speed of a point positioned at the edge of the disk at t=2s = 8 m/s