Answer to Question #86989 in Mechanics | Relativity for Maria

Question

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.

Answer 1

A.

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

A moment of inertia of yo-yo: 

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

From the  conservation of energy:

$$mgh = 0.5Iω^2 + 0.5mv^2$$

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

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

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

$$v=1.0 m/s$$

B.

$$\frac{E_r}{E_k}=\frac{0.5(\frac{I}{r^2})v^2}{0.5(\frac{I}{r^2}+ m)v^2}=\frac{1}{(\frac{mr^2}{I}+ 1)}$$

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