Kevin S. Huang

2007: Problem 15

Topic: System of Masses
Concepts: Pulleys with rotational inertia

Solution:

Applying Newton’s 2nd law to the mass, we have

mg-T=ma

Applying Newton’s 2nd law to the disk, we have

\tau=I\alpha

TR=\frac{1}{2}MR^2\alpha

Since there is no slipping, $a=R\alpha$ :

T=\frac{1}{2}Ma

Adding this equation to the first equation, we obtain

mg=\left(m+\frac{M}{2}\right)a

a=\frac{2mg}{2m+M}

The tension is then

T=\frac{1}{2}Ma=\frac{Mmg}{2m+M}=\frac{(8\,\mathrm{kg})(6\,\mathrm{kg})(10\,\mathrm{m/s^2})}{2(6\,\mathrm{kg})+8\,\mathrm{kg}}=24\,\mathrm{N}

so the answer is B.