A solid uniform disk has a string of negligible mass...

Question

A solid uniform 3.33 kg disk has string of negligible mass wrapped around its rim, with one end of the string tied to the ceiling, as shown in Figure 9.7. The disk is released from rest and turns as it falls as the string unwraps. At the instant its center has fallen 2.25 m

 

(a) how fast is it moving

 

(b) how much rotational kinetic energy does it have?

 

Thank you guys for your help

Answer 1

Use energy conservation. Potential energy transforms into rotational energy.

 

mgh=Iω²/2; Here I is the moment of inertia with respect to the axis passing through the point at the rim, where the rope leaves the disk. If you draw a picture, you will realize that the disk at every instant rotates around the point on the rim, just like the rolling wheel rotates around the point it touches the ground. The moment of inertia of the disk around its axis of symmetry, that is passing through the center, is I₀=mR²/2; the moment of inertia with respect to the shifted axis is I=I₀+md², where d is ωthe distance between two axes. In our case d=R. So we got:

 

mgh=(3/2)mR²ω²/2; From this equation we obtain:

 

ω=[√(4gh/3)]/R;

 

Disk's center of mass velocity is:

 

v=ωR=√(4gh/3); v=√[4*9.8 (m/s²)*2.25 (m)/3]≈5.42 m/s;

 

Its rotational energy is (rotation with respect to the center of mass):

 

K_r=I₀ω²/2=mR²/2*ω²/2=mgh/3; K_r=3.33 (kg)* 9.81 (m/s²) *2.25 (m)/3=24.5 J

Answer 2

   Treat this as a conservation of energy problem.  Potential energy is being converted into kinetic energy.  It may be helpful to treat the total kinetic energy  as comprised of translational  KE of the center of mass and the rotational KE about the center of mass.

 

   One can then write an equation     mgh = ½mv² + ½Iω²   where m is the mass (3.33 kg), and h is the distance the (center of) mass has fallen (2.25 meters).   v is the downward speed of the mass, I = ½mr² is the moment of inertia of the disc of mass m and radius r, and ω is the angular speed of the disc about its center of mass.  

 

  The key is to realize that v = rω.  (i.e. when a wheel or disc rolls without  slipping, the speed of the center depends on the  radius and angular speed about its center)

 

  Making this substitution for rω in I in the last term and combining the KE terms one has  mgh = ¾mv²   one can easily solve for v, and answer parts a) and b).

 

Comment:  Notice that v does not depend on either m or r, only on the distance that the center of mass drops.