# More Rotational Dynamics

1. Jan 12, 2006

### PhysicsDud

Ok I'm in desperate need of some direction! I have the following question in which I must answer twice once using rotational dynamics and then again using Conservation of Mechanical Energy.

Question:
A massless string is wrapped around a solid cylinder as
shown in the diagram at the right. A block of mass
kg 2.0 = m hangs from the string. When released, the
block falls a distance 82 cm in 2.0 s. Starting with a freebody
diagram, calculate the mass of the cylinder.

I have started with ma + 0.5m(cylinder)a= -mg
That got me a mass of 43.2 kg for the cylinder

determining a from the equation v^2 = v2^2 +2a (y-y2)
a = 0.1025 m/s^2

But I don't think that's right

Now for Mechanical Energy I'm completely lost.

2. Jan 12, 2006

### Gamma

Using S = ut + 1/2 a t^2, you can find acceleration a.

V = u + at should give you the final velocity.

For rotational part:

Use, Torque = I Alpha.

For mechanical energy:

Use conservation of energy :

Total energy of the system = potantial energy of the weight + Rotational KE of the cylinder + KE of the hanging mass.

3. Jan 13, 2006

### PhysicsDud

i don't understand what s and u represent?

Also, for mechanical energy I tried:
M=m1+m2 I = 0.5 m2 r^2 omega = v/r

0.5Mv1^2 + 0.5Iomega1^2 + Mgh1 =
0.5Mv2^2 + 0.5Iomega2^2 + Mgh2

E initial = m2gh = 2.0kg x 9.80 x 0 = 0

E final = m1gh + 0.5 m1v^2 + 0.5 Iomega^2 = m1gh + 0.5m1v^2 + 1/4 m2v^2

m2 = .425 ????????

What have I done wrong?

4. Jan 13, 2006

### Staff: Mentor

In Gamma's equation, s stands for distance and u stands for initial speed.

This is a bit confusing:
The cylinder (mass = m2) only rotates (I presume) so the only relevant energy for the cylinder is rotational KE (its gravitational PE doesn't change). The block (mass = m1), on the other hand, falls, so it has both translational KE and gravitational PE. So rewrite your mechanical energy equation.

5. Jan 13, 2006

### PhysicsDud

KE = KE + PE

0.5 m(cylinder) v^2 = 0.5 m(block) v^2 + m(block) gh

v = .82m/s
a = .41 m/s^2

Is this correct?

6. Jan 13, 2006

### Staff: Mentor

No. Think this way: Initially nothing is moving so the only energy is potential. As the block falls the potential energy decreases as the kinetic energy increases. The kinetic energy has two parts: The KE of the block ($$1/2 m_b v^2$$) plus the rotational KE of the cylinder ($$1/2 I_c \omega^2$$).

7. Jan 14, 2006

### PhysicsDud

Alright so let's see now, if there is only PE at first then the equation would be:

PE = KE(block) + KE(Cylinder)
mgy = $$1/2 m_b v^2$$ + $$1/2 I_c omega^2$$

Am I getting there?

8. Jan 14, 2006

### Gamma

yes, You are getting there.

9. Apr 27, 2008

### Lma12684

I am confused about the KE(cylinder) in the above equation. Can someone please explain this? Thanks.

10. Apr 30, 2008

### Staff: Mentor

Can you be more specific? What exactly do you not understand about it?

An object that rotates about its center of mass has a rotational KE equal to $1/2 I \omega^2$.