Answer:
The speed it reaches the bottom is
[tex]v=6.51m/s[/tex]
Explanation:
Given: [tex]m=5.0kg[/tex], [tex]r=47cm\frac{1m}{100cm}=0.47m[/tex]
Using the conservation of energy theorem
[tex]U_i=K_E+K_{ER}[/tex]
[tex]m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2[/tex]
[tex]v=r*w[/tex], [tex]I=\frac{1}{2}*m*r^2[/tex]
[tex]m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2[/tex]
[tex]m*g*h=\frac{3}{4}*m*r^2*w^2[/tex]
[tex]g*h=\frac{3}{4}*r^2*w^2[/tex]
Solve to w'
[tex]w^2=\frac{4*g*h}{3*r^2}[/tex]
[tex]h=x*sin(30)=6.5m*sin(30)=3.25m[/tex]
[tex]w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}[/tex]
[tex]w=27.74rad/s[/tex]
[tex]v=27.74rad/s*0.235m=6.51m/s[/tex]
The linear speed of the center of the disk when it reaches the bottom of the incline is 7.96 m/s.
The speed of the solid disk at the bottom of the incline is determined by applying principle of conservation of energy.
K.E = P.E
¹/₂mv² + ¹/₂Iω² = mgh
where;
¹/₂mv² + ¹/₂(¹/₂mr²)(v/r)² = mgh
¹/₂v² + ¹/₄v² = gh
6v² = 8gh
[tex]v = \sqrt{\frac{8gh}{6} }[/tex]
where;
[tex]v = \sqrt{\frac{8gLsin(\theta)}{6} } \\\\v = \sqrt{\frac{8\times 9.8 \times 6.5sin(40)}{6} } \\\\v = 7.96 \ m/s[/tex]
Thus, the linear speed of the center of the disk when it reaches the bottom of the incline is 7.96 m/s.
Learn more about moment of inertia of solid disk here: https://brainly.com/question/14531082
