# Archived Rotation with String Slipping & Not Slipping

1. Dec 2, 2013

### Morenzio

1. The problem statement, all variables and given/known data
Block's 1 (460g) & 2 (500g) are mounted on a horizontal axle of negligible friction (R = 5.00cm). When release from rest, block 2 falls 75.0 cm in 5.00s without the cord slipping on the pulley.

a) What is magnitude of acceleration of blocks?
b) Tension of T2
c) Tension of T1
d) What is magnitude of pulley's angular acceleration?
e) What is its rotational inertia?

Block 1 - 460g
Block 2 - 500g
Distance = 75.0 cm
Time = 5.00s

2. Relevant equations

x = x_0 + v_0 t + (1/2) a t^2

\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}

Can't seem to find other equation I was looking for on site involving (1/2)Mv2 + (1/2)Iω2

3. The attempt at a solution

a) The initial velocity is 0 m/s so plugging in givens into the first relevant equation results:
.75m = (1/2)a(5s)2
a=.06m/s2

b) After drawing a FBD, a is downwards thus:
F = ma
m2g - T2 = m2a
T2 = m2(g - a)
T2 = 4.87N

c) T1 - m1g = m1a
T1 = m1(g + a)
T1 = 4.54N

d) α = a/R = .06m/s2/.05m = 1.2 rad/s2

e) There is a loss of PE as block 2 falls and gain of PE as block 1 rises. The difference can be found:

m2gd = (.5kg)(9.8m/s2)(.75m) = 3.675 J
m1gd = (.46kg)(9.8m/s2)(.75m) = 3.381
Total PE Lost = .294 J

This can be equated to the final relevant equation (well I just assumed I probably am wrong) giving:

.294 J = .5Mv2 + .5Iω2
.294 J = .5(m1+m2)v2 + .5Iω2

However, we need v and we need ω:

v = at = (.06m/s2)(5s) = .3 m/s
ω = v/r = (.3 m/s)/(.05m) = 6 rad/s

Plugging into where I left off:

.294 J = .0432 kgm2/s2 + I(36 rad/s)
.2508 = 18I

Rotational inertia, I = .0139 kgm2

Problems I'm Having
If everything I did above is correct, I'm having trouble setting up this exact same problem if the string slips and the disk does not move. I would need to calculate the distance block 2 falls in 5 seconds. I was just assuming I was over thinking the problem and just think of it as free falling?

2. Feb 5, 2016

### haruspex

If the inertia of the pulley is to be ignored, there is a net force (difference in the weights) and a total mass (all undergoing the same acceleration). Or, more pedantically, set the two tensions equal.