Solutions to Rotational Dynamics Problems (3) - PART I

Solutions to the first four problems of Set (3):

 1(a) What unbalanced torque will produce an angular acceleration of 3 rad/sec²  in a 100 kg disk having a radius of gyration of 0.7 m?

Solution:  a  =  t / I and:  t  = aI

where angular acceleration:   a  =   3 rad/sec²

Since we are dealing with a disk its moment of inertia is given by:

I = ½ m r ²   = ½ (100 kg) (0.7 m) ²  =    24.5 kg · m²  

Unbalanced torque:  t  = aI  =   3 rad/sec² (24.5 kg · m² ) = 73.5 Nm

(b) Determine the mean effective torque that must be applied to a 40 kg flywheel with radius of gyration 0.2 m to give it an angular speed of 300 rpm in 10 secs?

Solution:  300 rpm = 10p rad/sec

angular acceleration developed in 10 sec:

a =(w ₁-w _o)/t  =  (10 p rad/s – 0)/ 10 s =   p   rad/sec²

Since we are dealing with a disk (flywheel) its moment of inertia is given by:

I = ½ m r ²   = ½ (40 kg) (0.2 m) ²  =    0.8 kg · m²  

The mean effective torque that must be applied:

t  = aI  = (p   rad/sec²) 0.8 kg · m²  =    2.51 Nm

3) For the example problem of the roller with friction force acting, assume a mass M = 1.5 kg and R = 0.15m.  If the angular acceleration is 1.0 rad/sec²   find:

a) The force F pulling the roller if the friction force = 0.

Soln:  F  -   F    = MA  =  ½ Mr ² a 

But:  F    = 0

So: F =  ½ Mr ² a   = ½  (1.5 kg) (0.15m) ²  1.0 rad/sec²

F = 0.016 N

b) The magnitude of the force  F  if the pulling force F = 15 N.

F    = - 1/3 F   =  -1/3 (15 N) = - 5N

c) The linear acceleration for case (a),  for case (b).

a) Soln.   F = MA,   A = F/M =  0.016 N/ (1.5 kg) = 0. 01 m/sec²

b) Soln. F  -   F    = MA  

A = (F  -  F ) /M = (15N -  (-5N)) / 1.5 kg  = 20 N/ 1.5 kg 

= 13. 3 m/sec²

A Yo-yo is made from a uniform disk of total mass M and radius R with a string wrapped around it as shown below:

The top end of the string is tied to the ceiling and the Yo-yo is released from rest with the string vertical.

a) Draw a force diagram showing the forces acting.

b) Write the equations for the translational and rotational motions.

i) Translational: Mg - T = MA

ii) Rotational:   RT = ½ MR ² a

c) Write equations for the angular acceleration and angular momentum.

a   =  2RT/MR ²      = 2 T/ MR

Finding the angular momentum L (= Iw)  requires doing part (d) first, since:

 w = v/R:

Whence: v =  Ö(4gh/ 3)    

So that: w =  Ö(4gh/ 3)/  R

So:  L =  I w  =  I  Ö(4g/ 3) h / R 

=  (½ MR² )Ö(4gh/ 3)/ R

L =  (½ MR)Ö(4gh/ 3)

d) Find the velocity of the Yo-yo after it has dropped a height h.

Go back to part (b):

Mg - T = MA

RT = ½ MR ² a  =  ½ MRA

(where we used the fact that A = Ra,  given the string unwinds at a rate proportional to the angular velocity  w  as the disk falls)

Now, divide the 2nd of the above eqns. by R after substituting the resulting equation for T into the first, i.e.

Mg - ½ MA  = MA  or:    

Mg =   MA + MA/2  =  3MA/2 

  so:   g =  3A/2

Or:  A = 2g/ 3   And:

  T =  ½ M (2g/3)  =  1/3 Mg

If the yo-yo starts from rest, the speed v of the center of mass after it has dropped a distance h is:

v =  Ö(2Ah)    since the acceleration is constant. 

 Substituting the value A = 2g/ 3:

v =  Ö(2(2g/ 3) h) =  Ö(4gh/ 3)