Solutions To Rotational Dynamics Problems (2)

1) A 1.0 kg hollow sphere with a radius of 10.0 cm rolls without slipping.  If the linear speed is p/2 m/s find: a) the translational kinetic energy, b) the rotational kinetic energy and c) the total kinetic energy of the hollow sphere.  (The moment of inertia for a hollow sphere is: I =   2/3  m R ²  )

Solution:

a) The translational kinetic energy, 

We have: K =  ½Mv ²

⁼    ½ (1.0 kg) (p/2 m/s)²   =    1.23 J

b) The rotational kinetic energy:

We have:

    K  =  ½ Iw ²  = ½ (½ Mr ² ) (v/r )²  = ½ (½ Mv ² )

= 0.61 J

c) The total kinetic energy of the hollow sphereL

This is simply the sum of parts (a) and (b):

K =  ½ I (v/r )²   +   ½Mv ²=  1.23 J + 0.61  J = 1.84 J

2) A 0.5 kg solid sphere with a radius of 5.0 cm rolls without slipping down an incline of height h = 0.5m.  Find: a) the initial total energy (kinetic and potential), and b) the final total energy, and c) the velocity at the bottom of the incline. (The moment of inertia for a solid sphere is: I  =   2/5   m R ² )

Solution:

a)    Initial total energy: K _i  +  V _i = 0 + mgh = mgh 

Where m = 0.5 kg,  R = 5.0 cm =  0.05m, h = 0.5m

Then: mgh = (o.5 kg) (9.8 m/sec²  ) (0.5 m) = 1.22 J

b)    the final total energy:

K _f  +  V _f = ½ mv ²  (1  +   I/mr ² )

I = 2/5   m R ² =   5 x 10 ^-4  kg · m²

c)     the velocity at the bottom of the incline.

v =    Ö2gh/ (1  +   I/mr ² ) =  2.64 m/s

So the final answer to part (b) is:

K _f  +  V _f =

 ½{0.5g(2.64 m/s)² (1  +   5 x 10 ^-4  kg · m²/ o.5 kg ( 0.05m) ² }

= 2.45 J

 

3)  A 1.0 kg solid ball rolling on a horizontal surface at 20 m/s  arrives at the bottom of an inclined plane which makes an angle of 30 degrees with the horizontal.  Find: a) The total kinetic energy of the ball when it has just arrived at the bottom of the incline, b) The distance up the incline the ball will continue to roll. (Ignore frictional forces).

Solutions:

a)    The total kinetic energy of the ball when it has just arrived at the bottom of the incline

K =  ½ I (v/r )²   +   ½mv ²= = ½ (2/5 m r ² ) (v/r )²   +   ½mv ²

7/10 mv ²  =   7/10 [1.0 kg) (20 m/s) ² ]  = 280 J

  b) The distance up the incline the ball will continue to roll.

Total KE at bottom =  PE at top

280 J  =  mgh   so  h =  280 J/  mg = 

280 J/ ([1.0 kg) (9.8 m/sec²  )  =  28.6 m

Distance moved up incline, call it ℓ:

ℓ  =  h/ sin 30 =  28.6m / 0.5 = 57.2 m

4) A wheel of radius 0.5 m and mass 5.0 kg rolls down a smooth incline 15 m high.  Find the wheel's linear speed when it is at the bottom of the incline. (Take the moment of inertia as I =   m R ² )                                        

Solution:

v =    Ö2gh/ (1  +   I/mr ² )

h = 15 m,   m= 5.0 kg,  g = 9.8 m/sec² 

I =   m R ²  =  1.25 kg · m²

Ö2gh/ (1  +   I/mr ² ) = 

Ö{2( 9.8 m/sec²  ) 15m / (1  +  1.25 kg · m² /5.0 kg (0.5 m) ² }     = 12.1 m/s