# Rotational Motion of two blocks

1. Oct 29, 2007

### Destrio

Two identical blocks, each of mass M, are connected by a light string over a frictionless pulley of radius R and rotational inertia I. One block is on a horizontal plane, one is dangling at the end of the string. The string does not slip on the pulley, and it is not known whether or not there is friction between the plane and the sliding block. when this system is released, it is found that the pulley turns through an angle (theta) in time t and the acceleration of the blocks is constant.

a) what is the angular acceleration of the pulley?
b) what is the acceleration of the two blocks?
c) what are the tensions in the upper and lower sections of the string?
All answers to be expressed in terms of M, I, R, (theta), g and t.

By biggest problem with this question is dealing with the pulley moving at an angle during the motion.

for the dangling block
in the y direction we have the force of gravity and the tension
and no force in the x

for the block on the platform
in the y direction we have no force
in the x direction we have the force of friction and the tension force

the 2 tension forces will be equal so I can relate the 2 blocks

how can I start dealing with the pulley though?
anyone have any hints to help me get started?

thanks

2. Oct 29, 2007

### learningphysics

It's just like kinematics... except with angles...

in kinemtics... an object starting from rest:

d = (1/2)at^2

But here:

theta = (1/2)alpha*t^2

solve for alpha.

3. Oct 30, 2007

### Destrio

α = 2Θ/t^2

Atangential = αR
at some point Atangential = a
a = αR

so
αR = 2Θ/t^2
α = 2ΘR/t^2

ΣFy = mg - T1 = ma
ΣFx = T2 - f = ma

T1 = T2

mg - T1 = T2 - f
mg + f = 2T
T = (mg+f)/2

Is this correct?

Thanks

4. Oct 30, 2007

### learningphysics

I'm confused by this...

α = 2Θ/t^2

a = αR

so a = 2ΘR/t^2

is that what you meant?

T1 is not necessarily equation to T2 in this situation, because the pulley has rotational inertia (ie it is not a massless pulley).

this part is right:

Mg - T1 = Ma

you can use the acceleration you derived before, to solve for T1 (vertical tension).

To get T2 (horizontal tension), use the torque equation for the pulley... you don't know friction, so you can't use that.

5. Nov 1, 2007

### Destrio

sorry yes

α = 2Θ/t^2

Atangential = αR
at some point Atangential = a
a = αR

so
αR = 2Θ/t^2
a = 2ΘR/t^2

ΣFy = Mg - T1 = Ma
T1 = Mg-Ma
T1 = M(g-a)
T1 = M[g-(2ΘR/t^2)]

τ = Iα
ΣFx = T2 - τ = ma
T2 = ma + τ
T2 = ma + Iα
T2 = m(2ΘR/t^2) + I(2Θ/t^2)
T2 = (2Θ/t^2)(mR + I)

is this all correct?

Thanks