A hoop starts from rest at a height 3.0 m above the base of an inclined plane and rolls down under the influence 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.)
Ke at bottom= PE at top
1/2 I W^2+ 1/2 m w^2 r^2=mg^3
solve for w, then v= wr
To find the linear speed of the hoop's center of mass just as it leaves the incline and rolls onto a horizontal surface, we can use conservation of energy.
The initial potential energy of the hoop is given by mgh, where m is the mass of the hoop, g is the acceleration due to gravity, and h is the initial height of the hoop above the base of the incline.
As the hoop rolls down the incline, its potential energy is converted into kinetic energy and rotational kinetic energy. The kinetic energy of an object is given by 1/2 mv^2, where v is the linear speed of the object.
The rotational kinetic energy of the hoop is given by 1/2 Iω^2, where I is the moment of inertia of the hoop and ω is the angular velocity of the hoop.
Since the hoop is rolling without slipping, the linear speed v of the hoop's center of mass is related to the angular velocity ω by the equation v = rω, where r is the radius of the hoop.
At the top of the incline, the hoop's potential energy is entirely converted into rotational kinetic energy, so we can write:
mgh = 1/2 Iω^2
To solve for ω, we need to find the moment of inertia I of the hoop. The moment of inertia of a hoop rotating about its central axis is given by I = mr^2, where m is the mass of the hoop and r is the radius of the hoop.
Substituting I = mr^2 into the equation, we have:
mgh = 1/2 (mr^2) ω^2
Canceling out the mass m from both sides, we get:
gh = 1/2 r^2 ω^2
Solving for ω, we have:
ω^2 = (2gh) / r^2
Taking the square root of both sides, we have:
ω = sqrt(2gh) / r
Now, we can substitute ω = v / r into the equation to find the linear speed v:
v / r = sqrt(2gh) / r
Cross-multiplying, we have:
v = r * sqrt(2gh) / r
Canceling out the radius r, we get:
v = sqrt(2gh)
Therefore, the linear speed of the hoop's center of mass just as it leaves the incline and rolls onto a horizontal surface is given by the equation v = sqrt(2gh), where g is the acceleration due to gravity and h is the initial height of the hoop above the base of the incline.