2015: Problems 14-15

Topic: Dynamics
Concepts: Circular motion, Newton’s laws


Solution:

Looking at one of the masses, we see that the tension provides the centripetal acceleration,

T=mω2r=mω2L2=(5kg)(4s1)2(3m)2=120NT=m\omega^2r=\frac{m\omega^2L}{2}=\frac{(5\,\mathrm{kg})(4\,\mathrm{s^{-1}})^2(3\,\mathrm{m})}{2}=120\,\mathrm{N}

so the answer is C.

Topic: System of Masses
Concepts: Conservation of angular momentum, Work-energy theorem


Solution:

There is no net external torque on the system of two masses since the tensions pulling them inward cancel out. Thus, we have conservation of angular momentum. Choosing the center of the rod as our pivot point, the initial moment of inertia IiI_i is

Ii=2m(L2)2=mL22=(5kg)(3m)22=22.5kg.m2I_i=2m\left(\frac{L}{2}\right)^2=\frac{mL^2}{2}=\frac{(5\,\mathrm{kg})(3\,\mathrm{m})^2}{2}=22.5\,\mathrm{kg.m^2}

The final moment of inertia IfI_f is

If=2mr2=2(5kg)(0.5m)2=2.5kg.m2I_f=2mr^2=2(5\,\mathrm{kg})(0.5\,\mathrm{m})^2=2.5\,\mathrm{kg.m^2}

The angular momentum LL is

L=Li=Iiω=(22.5kg.m2)(4s1)=90kgm2sL=L_i=I_i\omega=(22.5\,\mathrm{kg.m^2})(4\,\mathrm{s^{-1}})=90\,\mathrm{\frac{kg\,m^2}{s}}

By the work-energy theorem and using the fact that angular momentum is conserved,

W=EfEi=L22IfL22Ii=L22(1If1Ii)=9022(12.5122.5)J=1440JW=E_f-E_i=\frac{L^2}{2I_f}-\frac{L^2}{2I_i}=\frac{L^2}{2}\left(\frac{1}{I_f}-\frac{1}{I_i}\right)=\frac{90^2}{2}\left(\frac{1}{2.5}-\frac{1}{22.5}\right)\,\mathrm{J}=1440\,\mathrm{J}

so the answer is D.

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