Solutions to Rotational Dynamics Problems (3) - PART 2

 

Solutions To Problems 5-8:

5)  A cylinder of radius R = 0.5 m has a moment of inertia: I = 2.0  kg · m² about an axis of rotation. A rope is tied around the perimeter of the cylinder and a constant tension of 10 N is maintained. 

a) Find the magnitude of the torque due to the 10 N force about the axis of rotation.

Soln. Sketch forces diagram:

Torque  t  =  Fr  =  TR =   10 N x 0.5 m =  5 N-m

b) Find the change in angular momentum of the cylinder in a time interval of 4 sec.

Soln.  Since  t  =  dL/dt   and  t  = const., we can write the change in L as:

DL =  t Dt,  or:

DL =  5 Nm (4 sec) =    20  kg · m²  / sec

 

c) If the cylinder starts from rest what is the angular speed after 4 seconds. 

 

Soln. 

Since L = I w   and:  D L =  L  

 w  =  L/ I  =   (20  kg · m²  / sec)/ 2.0  kg · m²  

w =  10 rad/ sec

 

6) A torque 0f  5 N· m   is used to turn a nut with a force F = 15 N.  Find the distance from the location of applied force to the nut.

Soln.  t  =  5 nm,  F =  15 N

Distance r =  t  / F  =   5 nm/ 15m =  1/3 m

 

7) A light rope wrapped around a disk -shaped pulley is pulled tangentially with a force of 0.53 N. Find the torque applied to the disk and the angular acceleration of the pulley given that its mass is 1.3 kg and its radius is 0.11m.

Soln.   Torque applied to the disk:

t  =  Fr   =  (0.53N) (0.11m) = 0.0058 Nm

Since the pulley is a disk its moment of inertia is given by:

I = ½ m r ²   = ½ (1.3 kg) (0.11m) ²    =    7.9 x 10 ^-3     kg · m²  

Then the angular acceleration of the pulley is:

a  =  t / I  = (5.8 x 10 ^-2 Nm)

 8) In 2006 the Earth took 0.840 s  more time to complete 365 revolutions than it did in the year 1906. Find the average angular acceleration of the Earth in this time.

Solution:

Angular acceleration is equal to

a =(w -  w _o)/t 

 In this case t= 100 yrs. = 365 (revs) x 24 h x 3600 s/h

Let the time to complete 365  revolutions = t  in 1906. But in 2006 that number would be: t + 0.840 s, i.e. taking longer.

Since each revolution is 2 p  radians.  365 revolutions will be an angular displacement (  q _o  ) of  2 p radians.  Then:

  q _o       = (365 x 2 p rad)

So the average angular speed in 1906 is:   w _o  =    q _o   / t

= (365 x 2 p rad)/ (365 (revs) x 24 h x 3600 s/h)

Similarly, the average angular speed in 2006 could be written as:

  w   =    q _o   / t+ 0.840

=  (365 x 2 p rad)/ [(365 (revs) x 24 h x 3600 s/h) + 0.840s)

 

The difference between the two will be equal to:

D w =   (w -  w _o)/t 

=  (365 x 2 p) /t  -  (365 x 2 p) /t  + 0.840 s

=  (365 x 2 p) · (t  + 0.840) - t/ t ( t  + 0.840)

=  (365 x 2 p) · (0.840)/ t ( t  + 0.840)

Since t >>> 0.840s  we can use the approximation:

( t  + 0.840)   »   t 

Whence:

D w » (365 x 2 p) · (0.840) / t ( t  + 0.840)

D w » (365 x 2 p) · (0.840) / t ²  

 Now fill in the actual time in the denominator:

365 (revs) x 24 h x 3600 s/h) =  3.15 x 10 ⁷  s

So:  t ²     =  (3.15 x 10 ⁷  s) ²   = 9.9 x  10 ¹⁴  s ²

Finally: D w » (365 x 2 p) ·  (0.840)/ (9.9 x  10 ¹⁴  s ² )

D w » 1.946 x  10 ^-12  rad s^-1  

Then the average change in angular velocity over a period of 100 yrs.

D w/ D t    » 1.946 x  10 ^-12  rad s^-1  / [(100 x (365 (revs) x 24 h x 3600 s/h)]

»  6.17 x  10 ^{- 22}  rad s^-2