Acircular disk of radius 10 cm has a constant angular acceleration of 1.0 rad/s2 ; at t = 0 its angular velocity is 2.0 rad/s. (a) determine the disk’s angular velocity at t = 5.0 s . (b) what is the angle it has rotated through during this time? (c) what is the tangential acceleration of a point on the disk at t = 5.0 s ?

(a) ω=  7 rad/s

(b)  θ= 22.5 rad

(c)  at = 10 cm/s²

Explanation:

The uniformly accelerated circular movement, also called uniformly varied circular movement is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

at = α*R, where R is the radius of the movement

We apply the equations of circular motion uniformly accelerated :

ω= ω₀ + α*t  Formula (1)

θ=  ω₀*t + (1/2)*α*t² Formula (2)

at = α*R Formula (3)

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 speed ( rad/s)

ω : final angular speed ( rad/s)

R : radius of the circular path (cm)

at:  tangential acceleration, (cm/s²)

Data:

R= 10 cm  : radius of the disk

t₀=0 , ω₀ = 2 rad/s

α= 1 rad/s2

t = 5 s

(a) Disk’s angular velocity at t = 5.0 s

We replace data in the formula (1)

ω= ω₀ + α*t

ω=  2 + ( 1 )*(5)

ω=  7 rad/s

(b) Angle that the disk has rotated in t = 5.0 s

We replace data in the formula (2)

θ=  ω₀*t + (1/2)*α*t²

θ=  (2)*(5)+ (1/2)*(1)*(5)²

θ=  10+ 12.5

θ= 22.5 rad

c) Tangential acceleration of a point on the disk at t = 5.0 s

We replace data in the formula (2)

at = α*R

at = (1)*(10)

at = 10 cm/s²

i think the answer is a, hope this is right, that would be my best guess.

0.55x745.7j and a calculator