Dual Pulley Problem: Acceleration of 10kg Mass

[Boltak]

Hi everyone...

I've been struggling with a physics problem for 2 days now... I'm really not sure what to do.

Problem -
A pair of 10.0 kg masses are suspended from (massless) strings wrapped around a dual pulley. The radius of the smaller shaft on the pulley is half the radius of the larger shaft on the pulley. If the total mass of the pulley is also 10.0kg, and the pulley is considered to be a uniform disk, what is the acceleration (magnitude and direction) of mass ([tex]m_{1}[/tex]) connected to the larger shaft?

Basically... one 10.0 kg mass is attached on the right side of the small shaft, while other 10.0kg mass is attached to the left side of the bigger shaft.

picture at http://members.cox.net/lorddreg/p9.jpg

I was assuming [tex]m_{1}[/tex] would most likely accelerate down.

This is what I have so far... I'm not sure if it is correct or not ...

For [tex]m_{1}[/tex] ...
[tex]\sum{F = m * a_{y} = -T_{1} + m_{1} * g}[/tex]

For [tex]m_{2}[/tex] ...
[tex]\sum{F = m * a_{y} = T_{2} - m_{2} * g}[/tex]

I would appreciate any help!

Thank you

 

[StephenPrivitera]

You must consider torques about the axis through the center.
The net external torque is
[tex]m_1R - m_2r=I_{net}\alpha[/tex]
so
[tex]\alpha=\frac{m_1R - m_2r}{I_{net}}[/tex]
where
[tex]I_{net}=\frac{1}{2}MR^2+\frac{1}{2}mr^2+m_1R^2+m_2r^2[/tex]
you know r=1/2R, so you should be able to find the relation between m and M (the mass of the individual pulleys)
also, a_1=αR and a_2=αr
so there you have it

 

[M98Ranger]

for posting your question and diagram. It's always great to see someone actively seeking help and trying to understand a difficult problem.

First, let's define some variables for easier understanding. Let's call the smaller mass m_1 and the larger mass m_2. Also, let's define the radius of the smaller shaft as r_1 and the radius of the larger shaft as r_2.

Now, let's look at the forces acting on each mass. For m_1, there is the tension force from the string pulling upwards and the weight force pulling downwards. For m_2, there is the tension force from the string pulling downwards and the weight force pulling upwards.

Using Newton's second law, we can write the equations of motion for each mass:

For m_1:
\sum{F = m_1 * a = T_1 - m_1 * g}

For m_2:
\sum{F = m_2 * a = -T_2 + m_2 * g}

Since the pulley is considered a uniform disk, we can also use the equation for the rotational motion of a rigid body:

\sum{\tau = I * \alpha}

Where \tau is the net torque, I is the moment of inertia, and \alpha is the angular acceleration. In this case, the net torque is equal to the tension force multiplied by the radius of the pulley. The moment of inertia for a uniform disk is given by:

I = \frac{1}{2} * m * r^2

Where m is the mass of the pulley and r is the radius of the pulley. Plugging in the values for the smaller and larger shafts, we get:

For m_1:
T_1 * r_1 = \frac{1}{2} * m_2 * r_2^2 * \alpha

For m_2:
T_2 * r_2 = -\frac{1}{2} * m_2 * r_2^2 * \alpha

Solving for \alpha in both equations and setting them equal to each other, we get:

\alpha = \frac{2 * T_1 * r_1}{m_2 * r_2^2} = -\frac{2 * T_2 * r_2}{m_2 * r_2^2}

Solving for T_1 and T_2 in