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Zigmanuir [339]
3 years ago
13

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
Physics
1 answer:
saveliy_v [14]3 years ago
7 0

Answer:

The speed it reaches the bottom is

v=6.51m/s

Explanation:

Given: m=5.0kg, r=47cm\frac{1m}{100cm}=0.47m

Using the conservation of energy theorem

U_i=K_E+K_{ER}

m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2

v=r*w, I=\frac{1}{2}*m*r^2

m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2

m*g*h=\frac{3}{4}*m*r^2*w^2

g*h=\frac{3}{4}*r^2*w^2

Solve to w'

w^2=\frac{4*g*h}{3*r^2}

h=x*sin(30)=6.5m*sin(30)=3.25m

w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}

w=27.74rad/s

v=27.74rad/s*0.235m=6.51m/s

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