Answer:
h = 16.67m
Explanation:
If the kinetic energy of the cylinder is 510J:
[tex]Kc=510=1/2*Ic*\omega c^2[/tex]
[tex]\omega c=\sqrt{510*2/Ic}[/tex]
Where the inertia is given by:
[tex]Ic=1/2*m_c*R_c^2=1/2*(8)*(0.15)^2=0.0225kg.m^2[/tex]
Replacing this value:
[tex]\omega c=106.46rad/s[/tex]
Speed of the block will therefore be:
[tex]V_b=\omega_c*R_c=106.46*0.15=15.969m/s[/tex]
By conservation of energy:
Eo = Ef
Eo = 0
[tex]Ef = 510+1/2*m_b*V_b^2-m_b*g*h[/tex]
So,
[tex]0 = 510+1/2*m_b*V_b^2-m_b*g*h[/tex]
Solving for h we get:
h=16.67m
The mass would have to descend from a height of 13.01 meters.
Given the following data:
First of all, we would determine the moment of inertia for the solid cylinder by using this formula:
[tex]I=\frac{1}{2} mr^2\\\\I=\frac{1}{2} \times 8 \times 0.15^2\\\\I=4 \times 0.0225[/tex]
I = 0.09 [tex]Kgm^2[/tex]
Next, we would determine its angular velocity by using this formula:
[tex]K.E =\frac{1}{2} I\omega^2\\\\\omega=\sqrt{\frac{2K.E}{I} } \\\\\omega=\sqrt{\frac{2 \times 510}{0.09} }\\\\\omega=\sqrt{11,333.33} \\\\\omega=106.46\;rad/s.[/tex]
For the speed:
[tex]V=r \omega\\\\V= 0.15 \times 106.46[/tex]
V = 15.97 m/s.
Now, we would calculate the height by applying the law of conservation of energy:
[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\2gh=v^2\\\\h=\frac{v^2}{2g} \\\\h=\frac{15.97^2}{2\times 9.8} \\\\h=\frac{255}{19.6}[/tex]
h = 13.01 meters.
Read more on moment of inertia here: https://brainly.com/question/3406242
