Solutions To Basic Rotational Dynamics Problems (1)

 (1):An old-fashioned record turntable in the form of a uniform disk of 1.0 kg mass and radius R = 0.1m turns freely with an angular velocity of 8 rad/sec (about 78 rpm).  A dead bumble bee (from bug spray) lands at a point halfway between the center and the outside edge. If the mass of the dead bee is 0.1 kg, find the final angular velocity.

Solution:

r1=0.1 m

r2 = 0.1m/2 = 0.05m

w = 8 rad/s

M (turntable mass)= 1.0 kg   m (bee mass)  = 0.1kg

We know the rotational kinetic energy: K  = ½ Iw ²   =   ½Mw ² r ²

And: K before = K after

 

½Mw ² r1  = ½M w ² r2 ² + (M+m) w ² r2 ²

 

K (before bee drop) = ½ (½Mw ² r ² )  =

¼(1.0 kg) (8 rad/s) ²  (0.1m) ²

=0.16 J

K (after bee drop)=  {½M r2 ²  +  ½(M+m)  r2 ² }

Where (M + m) = 1.0 kg + 0.1 kg =  1.01 kg

Therefore:

½M r2 ²  =  ½ {(1.0 kg (0.1m) ²} = 0.005 kg · m²

And 2^nd term:

½(M+m)  r2 ²  = ½ {(1.01 kg (0.05m) ²} =  1.26 x 10 ^-3   kg · m²

But  w_f    is unknown and must be solved for:

Set Kinetic energy of rotation before = Kinetic energy of rotation after

 And solve for w_f:

w_f   =    Ö ½Mw ² r1 ²  /  {½M r2 ²  +  ½(M+m)  r2 ² }

w_f   =    Ö [0.16J  /  {(0.005 kg · m²) +  1.26 x 10 ^-3   kg · m² }]

 

w_f   =    5.05 rad/s

 

2) A flywheel is speeded up uniformly to 900 rpm (30p rad/sec) in 15 secs from rest. Find the angular acceleration a  in rad/sec² .

Solution:

a =(w ₁-w _o)/t  =  (30 p rad/s – 0)/ 15 s =  2 p   rad/sec²

 

3) A disc revolves with a constant acceleration of  5 rad/sec² .  How many turns does it make in 8 seconds from rest?

Solution:

The number of turns must refer to an angle q so that from rest:

q = w_o t  +   ½ at²  

Where:  a = 5 rad/sec²   w_o  = 0   and  t = 8s

Then:

q = ½  (5 rad/sec²) (8s) ²

q = ½  (5 rad x 64) = 160 rad

Or:   160 rad/  2 p   rad/rev = 25.5 revolutions (turns)

 

4) A compact disc has an angular speed of 210 rpm. Find it angular velocity in rad/s.

Solution:

We know in 1 revolution we get :  2 p   rad

And the angular speed is needed in rad/s, so we need to divide

210 rev/min x  2 p   rad/  (60 s/ min)  =  22 rad/s

 

5) A compact disc (CD) speeds up uniformly from rest to 310 rpm in 3.3 s.  How would you calculate the number of revolutions the CD makes in this time? How many would that be?

Solution:

Following an analogous procedure to Problem (4):

310 rpm = 310 rev/min x  2 p   rad/  (60 s/ min)  =  32.4 rad/s

The number of revolutions in 3.3 s must refer to an angle q  which we know started from rest ( w _o  = 0 rad/s)

q = w_o t  +   ½ at²  

First we need to obtain a :

a =(w ₁-  w _o)/t  =  (32.4 rad/s – 0)/ 3.3 s =  9.8   rad/sec²

 

q = ½ at²    =  ½ (9.8   rad/sec² )(3.3 s) ²

q = 53.3 rad

Or:   53.3 rad/  2 p   rad/rev = 8.4 revolutions

 

6) A massive uniform disk (e.g. grindstone) of radius R = 0.610 m has a moment of inertia I = 4.3 kg · m².  Find the mass of the grindstone.

Solution:

The moment of inertia for a uniform disk is: I =  ½ Mr ² 

Solving for the mass M:

M = 2 I/ r ²  =  2 ( 4.3 kg · m² )/ ( 0.610 m) ²   = 23.1 kg

 

7) A flywheel of mass 1.0 kg and radius 0.1m experiences a sudden change in angular velocity (from rest, i.e. w_i = 0 rad/s)  to  w_f = 20 rad/sec.  Has an external torque been applied to cause this?

Solution:

To check for an external torque we need to see if the following condition for angular momentum applies:

I_i(w_i) = I_f (w_f )= const.  

For a flywheel – given it’s a uniform disk- the moment of inertia is:

I =  ½ Mr ²    

I =  ½ (1.0 kg (0.1m) ²) = 0.005 kg · m²

Now,  w_i = 0 rad/s  so I_i(w_i) = 0

But I_f (w_f )=  (0.005 kg · m² )( 20 rad/sec) = 0.1 kg · m² /s

So  an external torque is concluded to apply