Question

A flywheel is a solid, rotating disk that is sometimes used to store energy. The moment of inertia for a solid disk is 1/2MR^2 where M is the mass and R is the radius. If a flywheel weighs 15kg and has a radius of 0.8m, determine the following.
  
A) If a point on the outside of the disk is moving at 6 m/s, what is the angular velocity of the flywheel?
  
B) If the flywheel starts from rest and accelerates at a constant rate for 60 s to bring it up to the angular velocity found in Part A, what is the angular acceleration?
  
C) If the angular acceleration in Part B is generated by a constant force applied tangentially to the outside edge of the disk (as shown below), what is the magnitude of the force?
  

Answer 1

Part A)

w = v/r

w = (6)/(.8)

w = 7.5 rad/s

Part B)
Apply wf = wo + (alpha)t

7.5 = 0 + alpha(60)

alpha = .125 rad/s^2

Part C)
Apply Torque = I(alpha) and FL, Thus

(.5mr^2)(.125) = F(r)

.5(15)(.8)(.125) = F

F = .75 N

Answer 2

a)

angular velocity, w = v / r

= 6 / 0.8

= 7.5 rad./ s

b)

Angular acceleration, alpha = (7.5 - 0) / 60

= 0.125 rad./s^2.

c)

F = torque / r

= I x alpha / r (where, I = moment of Inertia of disk.)

= (1/2) x m x r^2 x 0.125 / r

= 0.5 x 15 x 0.8 x 0.125

= 0.75 N.

Answer 3

A)omega= V/R=6/0.8=7.5 rad/s

B)angular cceleration=7.5/60=0.125 rad/s^2

C)T=I*Alfa

I=0.5*M*R^2=0.5*15*(0.8)^2=4.8 kg m^2

T=0.125*4.8=0.6 N-m

T=F*R

F=T/R=0.6/0.8=0.75 N ans

Answer 4

A) v= Rw, w = 6/.8 =7.5 rad./s

B) because w = w0+at so a = w/t = .125rad/s^2

C)because F*R = I*a so F = I*a/R =.75n