Question

1. A flywheel is initially rotating at 30 rad/s and has a constant angular acceleration. After 8.0 s it has rotated through 460 rad. Its angular acceleration in rad/s is:

2.Ten seconds after an electric fan is turned on, the fan rotates at 6000 rev/min. Its average angular acceleration in rad/s² is:

3.The rotational inertia of a disk about its axis is 0.70 kgˑm². When a 2.0-kg weight is added to its rim, 0.50 m from the axis, the rotational inertia in kgˑm² becomes:

4.A disk with a rotational inertia of 5.0 kgˑm² and a radius of 0.25 m rotates on a frictionless fixed axis perpendicular to the disk and through its center. A force of 8.0 N is applied along
the rim. The angular acceleration of the disk in rad/s² is:

5.A disk has a rotational inertia of 6.0 kgˑm² and a constant angular acceleration of 2.0 rad/s². If it starts from rest the work (in J) done during the first 6.0 s by the net torque acting on it is:

Answer 1

Solution) (1) wo = 30 rad/s

t = 8 s

Theeta = 460 rad

Angular acceleration = ?

Theeta = (wo)t + (1/2)(angular acceleration)(t^2)

460 = (30×8) + (1/2)(angular acceleration)(8^2)

220 = (32)(angular acceleration)

Angular acceleration = 220/32 = 6.875 rad/s^2

Angular acceleration = 6.875 rad/s^2

(2) t = 10 s

w = 6000 rev/min

1 rev/min = (2(pi))/(60) rad/s

w = 6000×(2(pi))/(60)

w = 628.31 rad/s

Average angular acceleration = w/t = 628.31/10

Average angular acceleration = 62.83 rad/s^2

(3) I = 0.70 kgm^2

m = 2kg

r = 0.50 m

I' = ?

I' = I + mr^2

I' = 0.70 + (2)(0.50^2)

I' = 1.2 kgm^2

(4) I = 5 kgm^2

r = 0.25 m

F = 8 N

Angular acceleration = ?

We have torque T = I(angular acceleration)

T = r × F

r × F = I(angular acceleration)

Angular acceleration = ( r × F)/(I) = (0.25×8)/(5)

Angular acceleration = 0.4 rad/s^2

(5) I = 6 kgm^2

Angular acceleration = 2 rad/s^2

wo = 0 (rest)

Time t = 6 s

Work W = T(theeta)

Angular displacement theeta

Theeta= (wo)t + (1/2)(angular acceleration)t^2

Theeta = 0 + (1/2)(2)(6^2)

Theeta = 36 rad

T = I(angular acceleration)

T = 6×2 = 12 Nm

W = T(theeta)

W = 12×36 = 432 J

W = 432 J