Respuesta :
Answer:
The velocities of hoop , disk and sphere are 1.89 m/s , 2.18 m/s , 2.26 m/s.
Explanation:
Lets find the speed of any general body of mass 'm' , moment of inertia 'I' , radius 'r'.
Let 'v' be the speed and 'ω' be the angular velocity of the body at bottom of the slope.
Since there is no external force acting on the system (Eventhough friction is acting at the point of contact of the body and slope , it does no work as the point of contact is always at rest and not moving) , we can conserve energy for this system.
Initially the body is at rest and at a vertical height 'h' from the ground.
Here , h=3sin(7°)
Initial energy = mgh.
Finally on reaching bottom h=0 but the body has both rotational and translational kinetic energy.
∴ Final energy = [tex]\dfrac{1 }{2}[/tex]I[tex]ω^{2}[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex].
Since the body is rolling without slipping.
v=rω
and
Initial Energy = Final Energy
mgh = [tex]\dfrac{1 }{2}[/tex]I[tex]ω^{2}[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex]
∴ mgh = [tex]\dfrac{1 }{2}[/tex]I[tex]\dfrac{v^{2} }{r^{2} }[/tex] + [tex]\dfrac{1 }{2}[/tex]m[tex]v^{2}[/tex]
∴ v = [tex]\sqrt{\frac{2mgh}{\frac{I}{r^{2} }+m } }[/tex]
For a hoop ,
I = m[tex]r^{2}[/tex]
Substituting above value of I in the expression of v.
We get,
v = [tex]\sqrt{gh}[/tex] = [tex]\sqrt{9.81×3sin(7°) }[/tex] = 1.89 m/s
Similarly for disk,
I = [tex]\dfrac{1}{2}[/tex]m[tex]r^{2}[/tex]
We get,
v = [tex]\sqrt{\frac{4gh}{3} }[/tex] = 2.18 m/s
For solid sphere ,
I = [tex]\dfrac{2}{5}[/tex]m[tex]r^{2}[/tex]
v = [tex]\sqrt{\frac{10gh}{7} }[/tex] = 2.26 m/s.
