More Rotational Dynamics

[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.

 

[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.

 

[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?

 

[Doc Al]

i don't understand what s and u represent?

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

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

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.

 

[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?

 

[Doc Al]

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 ($$1/2 I_c \omega^2$$).

 

[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?

 

[Gamma]

yes, You are getting there.

 

[Lma12684]

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

 

[Doc Al]

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

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 $1/2 I \omega^2$.