Conservation of energy and string pulley problem

  • Thread starter gunster
  • Start date
  • #1
7
0

Homework Statement


Consider two masses of 3.3 kg and 7.4 kg
connected by a string passing over a pulley
having a moment of inertia 12 g · m2
about its axis of rotation, as in the figure below. The
string does not slip on the pulley, and the
system is released from rest. The radius of
the pulley is 0.35 m.

Find the linear speed of the masses after
the 7.4 kg mass descends through a distance
21 cm. Assume mechanical energy is conserved during the motion. The acceleration of
gravity is 9.8 m/s^2.

Answer in units of m/s

Homework Equations



delta PE = KE
KE = K(translational) + K(rotational)

The Attempt at a Solution



Found that change in potential energy should be equal to change in potential energy of the heavier mass (where PE is lost) subtracted by the change in potential energy of the lighter mass (where some PE is gained).

Therefore: 7.4*g*(21/100) - 3.3 * g * (21/100) = delta PE

I then set total change in KE to the delta PE. I determined rotational KE to be 1/2 I * omega^2

where I is given to me and omega is (v/r)^2 and r is given to me.

K translational = 1/2 * Mtotal * v^2. I factored out V^2, and set that equal to delta PE / rest of that mess

So in the end: v = sqrt ( delta PE / (6/r^2 + 1/2M)).

However, it wasn't the right answer :( so any ideas?
 

Answers and Replies

  • #2
gneill
Mentor
20,913
2,862
In what units is the moment of inertia given?
 
  • #3
7
0
wow I see the moment of inertia is given in grams. Thank you so much that would explain why I failed XD
 

Related Threads on Conservation of energy and string pulley problem

Replies
24
Views
3K
  • Last Post
Replies
1
Views
2K
Replies
6
Views
2K
Replies
15
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
8
Views
3K
Replies
1
Views
4K
Replies
24
Views
12K
Top