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A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that is 6.50 m long. g = 9.81 m/s2 . (a) Calculate the linear speed of the center of the disk when it reaches the bottom of the incline. nothing m/s

Respuesta :

davidlcastiblanco

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]

onyebuchinnaji

The linear speed of the center of the disk when it reaches the bottom of the incline is 7.96 m/s.

Speed of the disk at the bottom of the incline

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;

  • I is moment of inertia of the uniform solid disk = ¹/₂mr²
  • ω is angular speed = v/r

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;

  • h is the height of the incline, h = Lsinθ

[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