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Archy [21]
3 years ago
8

A solid cylinder of mass 3.0 kg and radius 0.2 m starts from rest at the top of a ramp, inclined 15°, and rolls to the bottom wi

thout slipping. (For a cylinder I = MR2.) The upper end of the ramp is 1.5 m higher than the lower end. Find the linear speed of the cylinder when it reaches the bottom of the ramp. (g = 9.8 m/s2)
Physics
1 answer:
Lera25 [3.4K]3 years ago
3 0

Answer:

v = 4.4271 m/s

Explanation:

Given

m = 3 Kg

R = 0.2 m

∅ = 15°

h = 1.5 m

g = 9.8 m/s²

v = ?

Ignoring frictional losses, at the bottom of the plane

Total kinetic energy is  =  Potential Energy at the top of plane

Using Law of conservation of energy we have

U = Kt + Kr

m*g*h = 0.5*m*v² + 0.5*I*ω²

knowing that

Icylinder = 0.5*m*R²

ω = v/R

we have

m*g*h = 0.5*m*v² + 0.5*(0.5*mR²)*(v/R)² = 0.75*m*v²

⇒  v = √(g*h/0.75) = √(9.8 m/s²*1.5 m/0.75)

⇒  v = 4.4271 m/s

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Explanation:

Given :

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Angular velocity of the pulley is given by

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For the rotating body, we have

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KE_b=12J

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Answer:

v = 40 m / s

Explanation:

Let's use the expressions for accelerated motion

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where vo is the initial velocity, at the acceleration and t is the time.

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Answer:

I gonna give you the number so but you need to round 6.19047619048

Explanation:

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