Acceleration of Two Masses on a Pulley with Moment of Inertia

[Karol]

Homework Statement

The pulley on the drawing has moment of inertia of I=kMR².
What's the acceleration.

Homework Equations

$$M=I\alpha$$

$$F=ma$$

The Attempt at a Solution

$$m_1g-T_1=m_1a$$
$$T_2-m_2=m_2a$$
$$\mbox{ }R\alpha=a$$
$$(T_1-T_2)R=kMR^2\cdot\alpha$$

all these yield:
$$a=\frac{g(m_1+m_2)}{kM+m_1+m_2}$$
While the answer should be:
$$a=\frac{g(m_1-m_2)}{kM+m_1+m_2}$$

 

[ehild]

The Attempt at a Solution

$$m_1g-T_1=m_1a$$
$$T_2-m_2=m_2a$$
$$\mbox{ }R\alpha=a$$
$$(T_1-T_2)R=kMR^2\cdot\alpha$$

All is correct up to here, but the missing g in the second equation.

all these yield:
$$a=\frac{g(m_1+m_2)}{kM+m_1+m_2}$$

Check it.

T₁=m₁(g-a),
T₂=m₂(g+a)

T₁-T₂?ehild

 

[Karol]

Thanks!