Question #86989
A yo-yo is made of two solid cylindrical disks, each of mass 0.049 kg and diameter 0.070 m , joined by a (concentric) thin solid cylindrical hub of mass 0.0055 kg and diameter 0.012 m .
Part A
Use conservation of energy to calculate the linear speed of the yo-yo when it reaches the end of its 0.90 m long string, if it is released from rest.
Express your answer using three significant figures and include the appropriate units.
Part B
What fraction of its kinetic energy is rotational?
Express your answer using two significant figures.
1
Expert's answer
2019-04-02T10:46:19-0400

A.


m=2(0.049)+0.0055=0.1035kgm=2(0.049)+0.0055=0.1035kg

A moment of inertia of yo-yo: 


I=2(0.5)(0.049)(0.035)2+(0.5)(0.0055)(0.006)2=0.00006kgm2I=2(0.5)(0.049)(0.035)^2 + (0.5)(0.0055)(0.006)^2=0.00006 kgm^2

From the  conservation of energy:


mgh=0.5Iω2+0.5mv2mgh = 0.5Iω^2 + 0.5mv^2

mgh=0.5(I)(vr)2+0.5mv2mgh =0.5(I)(\frac{v}{r})^2 +0.5mv^2

mgh=0.5(Ir2+m)v2mgh = 0.5(\frac{I}{r^2}+ m)v^2

(0.1035)(9.8)(0.9)=0.5(0.000060.0062+0.1035)v2(0.1035)(9.8)(0.9)= 0.5(\frac{0.00006}{0.006^2}+ 0.1035)v^2

v=1.0m/sv=1.0 m/s


B.


ErEk=0.5(Ir2)v20.5(Ir2+m)v2=1(mr2I+1)\frac{E_r}{E_k}=\frac{0.5(\frac{I}{r^2})v^2}{0.5(\frac{I}{r^2}+ m)v^2}=\frac{1}{(\frac{mr^2}{I}+ 1)}

ErEk=1((0.1035)0.00620.00006+1)=0.94\frac{E_r}{E_k}=\frac{1}{(\frac{(0.1035)0.006^2}{0.00006}+ 1)}=0.94


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