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3 years ago

10

g A 5.60-kilogram hoop starts from rest at a height 1.80 m above the base of an inclined plane and rolls down under the influenc

e of gravity. What is the linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface? (Neglect friction.)

2 answers:

noname [10]3 years ago

6 0

Answer:

Linear speed = 4.2 m/s

Explanation:

Potential Energy at top = Kinetic Energy at bottom

Now, KE is the sum of linear and rotational kinetic energy.

Thus,

mgh = (1/2)mv² + (1/2)Iω²

Moment of Inertia for hoop; I = mr² Also, angular velocity; ω = v/r

Thus,

mgh = (1/2)mv² + (1/2)(mr²)(v/r)² = mgh = (1/2)mv² + (1/2)mv²

mgh = mv²

The m will cancel out and so;

gh = v²

9.81(1.80) = v²

v² = 17.658

v = √17.658

v = 4.2 m/s

swat323 years ago

4 0

Answer:

4.24m/s

Explanation:

Potential energy at the top= kinetic energy at the button

But kinetic energy= sum of linear and rotational kinetic energy of the hoop

PE= mgh

KE= 1/2 mv^2

RE= 1/2 I ω^2

Where

m= mass of the hoop

v= linear velocity

g= acceleration due to gravity

h= height

I= moment of inertia

ω= angular velocity of the hoop.

But

I = m r^2 for hoop and ω = v/r

giving

m g h = 1/2 m v^2 + 1/2 (m r^2) (v^2/r^2) = 1/2 m v^2 + 1/2 m v^2 = m v^2

and m's cancel

g h = v^2

Hence

v= √gh

v= √10×1.8

v= 4.24m/s

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