**1. The problem statement, all variables and given/known data**

A block with mass m = 5.00kg slides down a surface inclined 36.9 ∘ to the horizontal (the figure (Figure 1) ). The coefficient of kinetic friction is 0.27. A string attached to the block is wrapped around a flywheel on a fixed axis at O. The flywheel has mass 23.0kg and moment of inertia 0.500 kg⋅m2 with respect to the axis of rotation. The string pulls without slipping at a perpendicular distance of 0.200m from that axis.

What is the acceleration of the block down the plane?

**2. Relevant equations**

Moment of Inertia

[itex]\tau = rF = I\alpha[/itex]

**3. The attempt at a solution**

R = radius of wheel = [itex] I = \frac{MR^2}{2} => \sqrt(\frac{1}{M})=R [/itex]

r = radius to point P (where the force acts as a torque)

[itex]F_{net x } = mgsin\theta-\mu_kmgcos\theta-T=ma[/itex]

[itex] \tau = rT = I\alpha = \frac{Ia}{R}[/itex]

[itex] T = \frac{Ia}{rR} [/itex]

[itex] mg(sin\theta – \mu_kcos\theta)=ma+\frac{Ia}{rR}[/itex]

[itex] mg(sin\theta – \mu_kcos\theta)\frac{1}{m+\frac{I}{rR}}=a[/itex]

Somehow this is giving me the wrong answer, what did I do wrong?

EDIT: it seems that when you set R=r it’ll give you the right answer. This does not make sense, the radius of the Cylinder is not equivalent to the radius of the torque (which is defined as .2 in the problem). They differ, if only slightly, however mastering physics tells me I’m wrong.

http://ift.tt/1g9YG33