Conservation Laws in Rotational Motion

[Draggu]

Homework Statement

15. A solid sphere of mass 6.0 kg is mounted on a vertical axis and can rotate freely without friction. A massless cord is wrapped around the middle of the sphere and passes over a 1.0 kg pulley and is attached to block of mass 4.0 kg, as shown. What is the speed of the block after it has fallen 80 cm? Treat the pulley as solid cylinder.

Homework Equations

E initial = E after
ωp = angular velocity of pulley
ωs = angular velocity of sphere
rp = radius of pulley
rs = radius of sphere

Forces
-------

Pulley: Iα = Rp(T2-T1)
(rp^2/2)(a/rp) = Rp(T2-T1)
=(a)/2 = T2-T1

Block: mg-T2 = ma
=39.2 - T2 = 4a

Sphere: Iα = Rs(T1)
T1 = (2/5)(6)(rs)^2
T1= 2a/5

a = 8m/s^2
T2=7.2N
T1=3.2N


v^2-v0^2 = 2ad
v0 = 0

v^2 = 2(8)(.8)
v=3.57m/s


So initially I thought energy would be used in this situation, but now I'm trying to think if it's even necessary.

 

[Simon Bridge]

There is more than one way to approach the problem fersure.
Just pick the one you feel comfy with.

 

[Draggu]

When I try with the energy conservation method, I get a different number. Perhaps you could offer some insight into this?

mgh = ΔKE block + ΔKE pulley + ΔKE sphere
mgh = (0.5)(4)(v₂²-v₁²) + (0.5)(I_pulley)(ω₂²-ω₁²) + (0.5)(I_sphere)(ω₂²-ω₁²)
=(0.5)(4)(v₂²) + (0.5)(0.5)(1kg)(R_p²)(ω₂²) + (0.5)(2/5)(6kg)(R_s²)(ω₂²)
(4kg)(9.8)(0.8m) = 2(v₂²) + (1/4)(R_p²)(v_p²/R_p²) + (6/5)(R_s²)(v_s²/R_s²)
31.36 = 2(v_b²) + (1/4)(v_p²) + (6/5)(v_s²)

Not too sure where to go from here.

 

[Simon Bridge]

relate v_p and v_s to the speed of the block.

 

[Draggu]

How would I go about doing that? Assuming the velocities are the same, I get 3.01m/s.. which is different than the value from forces. Hmm, so you say to relate them. Which means either vb = vp = vs, or vb = vp + vs?

 

[haruspex]

How would I go about doing that? Assuming the velocities are the same, I get 3.01m/s.. which is different than the value from forces. Hmm, so you say to relate them. Which means either vb = vp = vs, or vb = vp + vs?

vb = vp = vs, as you did to obtain the (correct) result in post #5.
In the OP, these equations are wrong:

(rp^2/2)(a/rp) = Rp(T2-T1)
=(a)/2 = T2-T1

Check the dimensions of each side.

 

[Draggu]

vb = vp = vs, as you did to obtain the (correct) result in post #5.
In the OP, these equations are wrong:

Check the dimensions of each side.

Hi, thanks for replying.

I re-did the question and figured out what went wrong. I'm extremely silly... in the net force for the sphere I wrote "2a/5" instead of "12a/5" heh. That was enough to mess everything up! Confirmed, got the same answers using both methods. Thanks guys.

 

[Simon Bridge]

Well done. :)

 

[Grabo]

Sorry, but this makes absolutely zero sense to me. Without radius, I am hooped. I have tried both the energy
and force approach and gotten stuck midway through.

Homework Statement

15. A solid sphere of mass 6.0 kg is mounted on a vertical axis and can rotate freely without friction. A massless cord is wrapped around the middle of the sphere and passes over a 1.0 kg pulley and is attached to block of mass 4.0 kg, as shown. What is the speed of the block after it has fallen 80 cm? Treat the pulley as solid cylinder.

Homework Equations

E initial = E after
ωp = angular velocity of pulley
ωs = angular velocity of sphere
rp = radius of pulley
rs = radius of sphere

Forces
-------

Pulley: Iα = Rp(T2-T1)
(rp^2/2)(a/rp) = Rp(T2-T1)
=(a)/2 = T2-T1

Block: mg-T2 = ma
=39.2 - T2 = 4a

Sphere: Iα = Rs(T1)
T1 = (2/5)(6)(rs)^2
T1= 2a/5

a = 8m/s^2
T2=7.2N
T1=3.2Nv^2-v0^2 = 2ad
v0 = 0

v^2 = 2(8)(.8)
v=3.57m/sSo initially I thought energy would be used in this situation, but now I'm trying to think if it's even necessary.

I have no clue how you solved for this...how did you go from Ia=r(sphere)(T1) to T1=(2/5)(6)r(sphere)^2. Shouldn't it go to T1=mr^2a/r(sphere). I am so lost. Could somebody just explain to me how you got the tensions + how to not get stuck since radius is not given. Please!

 

[haruspex]

Ia=r(sphere)(T1) to T1=(2/5)(6)r^2. Shouldn't it go to T1=mr^2a/r.

Why would it do that? What is I in terms of m and r for a solid sphere? Please post your working.