# Rotational motion problem

1. Mar 28, 2013

### bona0002

Hey guys,

So, I'm kinda distraught over this problem because, by all accounts, I should be able to get the problem no sweat. However, it just hasn't been happening. If you could point out my flaw, I would truly appreciate it!

In the problem, we are asked to find the magnitude of the acceleration of the two blocks. This is how I approached it:

For block m_1_, the forces acting upon it are:

ƩF_x_ = T_1_ - (μ_k_)(N_1_) = m_1_a (Obviously, because the frictional force opposes the motion of the tension.)
ƩF_y_ = N_1_ - m_1_g = 0

For block m_2_, the forces acting upon it are:

ƩF_x_ = -T_2_ - (μ_k_)(N_2_) + (m_2_)(g)(sin θ) = m_2_a
ƩF_y_ = N_2_ - (m_2_)(g)(cos θ) = 0

Finally, for the pulley:

Ʃtorques (couldn't find the symbol) = (I)(α)

Now, assuming that the rope doesn't slip, α = a/r. Also, the moment of inertia = (1/2)MR^2

With all this information in mind, the equation to solve for a:

T_1_R - T_2_R = (I)(α)
= (μ_k_*N_1_ + m_1_*a)(R) - (-μ_k_*N_2_ + m_2_*g*sin θ - m_2_*a)(R) = ((1/2)MR^2)(a/R)

On the left side of the equation, the R in the numerator can cancel with the R in the denominator. In addition, an R can be factored out of the equation on the left side and can cancel with the remaining R on the right side. Now, the equation becomes:

(μ_k_*N_1_) + (m_1_*a) + (μ_k_*N_2_) - (m_2_)(g)(sin θ) + (m_2_*a) = (1/2)Ma

= (m_1_*a) + (m_2_*a) - (1/2)Ma = - (μ_k_*N_1_) - (μ_k_*N_2_) + (m_2_)(g)(sin θ)

=a(m_1_ + m_2_ - (1/2)M) = - (μ_k_*N_1_) - (μ_k_*N_2_) + (m_2_)(g)(sin θ)

=a = (- (μ_k_*N_1_) - (μ_k_*N_2_) + (m_2_)(g)(sin θ)) / ((m_1_ + m_2_ - (1/2)M))

So, after plugging in all the values, I arrive at 7.65 m/s^2. What am I doing wrong?

Thanks for the help!

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2. Mar 28, 2013

### TSny

What did you choose for the positive direction of your torques and angular acceleration? Does your choice of direction for positive angular acceleration correspond with a positive value for the linear acceleration?