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**1. Homework Statement**

A 4 kg collar C rides along a horizontally rotating arm AB. A spring k connects C to A. The arm AB rotates at a constant rate of 15 rpm. At the illustrated instant r = 1 m, r˙ = 1.1 m/s, and r¨ = 4 m/s2. The coefficient of dynamic friction, μk, is equal to 0.3. What is the magnitude of the total force exerted by the spring on the collar?

http://img337.imageshack.us/img337/958/dynamicsqig9.th.gif [Broken]

**2. Homework Equations**

[tex]a=(\ddot {r} -r \dot {\theta}^2)i_R + (2 \dot {r} \dot {\theta} + r \ddot {\theta})i_ {\theta}[/tex]

**3. The Attempt at a Solution**

[tex]\\sumF_r=ma_r=m(\ddot {r} -r \dot {\theta}^2)=-F_s-\mu R[/tex]

[tex]\\sumF_{\theta}=ma_{\theta}=m(2 \dot {r} \dot {\theta} + r \ddot {\theta})=R[/tex]

Everything is known exepth for [tex]F_s[/tex] which is the answer (force on spring), R, and [tex]\dot {\theta}[/tex].

[tex]\dot {\theta}=\frac {rpm*2 \pi}{60} = 1.57[/tex]

I plug in numbers and get R=13.82 and Fs=-10.27

Any help would be appreciated

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