A uniform-density wheel of mass 9 kg and radius 0.40 m rotates on a low-friction axle. Starting from rest, a string wrapped around the edge exerts a constant force of 13 N for 0.72 s. (a) What is the final angular speed? Entry field with correct answer 5.2 radians/s (b) What was the average angular speed? Entry field with incorrect answer 1.872 radians/s (c) Through how big an angle did the wheel turn? Entry field with incorrect answer 1.3478 radians/s (d) How much string came off the wheel? Entry field with incorrect answer 0.8469 m

Respuesta :

Answer:

Explanation:

Moment of inertia of wheel = 1/2 x mR²  , m is mass and R is radius of wheel

= .5 x 9 x .4²

= .72 kg m²

Torque created on wheel by string = T x r , T is tension and r is radius of wheel .

13 x .4 = 5.2 N m

angular acceleration α = torque / moment of inertia

= 5.2 / .72

= 7.222 rad /s²

a ) final angular speed = α x t , α is angular acceleration , t is time.

=  7.222  x .72

= 5.2 rad /s

b )

θ = 1/2 α t² , θ  is angle turned , t is time

= .5 x 7.222 x .72²

= 1.872 rad

average angular speed = θ / t

= 1.872 / .72

= 2.6 rad /s

c )

angle turned = 1.872 rad ( discussed above )

d )

length of string coming off

=  angle rotated x radius of wheel

= 1.872 x .4

= .7488 m .

74.88 cm