The required ratio of L'/L is [tex]\sqrt{21/5}[/tex] .
Given data:
The mass of solid sphere is, m.
The radius of roll is, R.
The vertical dropping height is, H.
The horizontal distance covered by shell to hit the ground is, L.
The horizontal distance covered by sphere to hit the ground is, L'.
The given problem can be resolved using the concept of conservation of energy.
The Potential energy is,
U = mgD
Here, D is the distance from sliping point to ground.
And Kinetic energy is,
[tex]KE=\dfrac{1}{2}mv^{2}+\dfrac{1}{2}I \omega^{2}[/tex]
Here,
I is the moment of inertia of spherical shell.
v is the linear speed of shell.
Apply the conservation of energy as,
potential energy = kinetic energy
U = KE
[tex]U = KE\\\\mgD=\dfrac{1}{2}mv^{2}+\dfrac{1}{2}I \omega^{2}\\\\\\mgD=\dfrac{1}{2}mv^{2}+\dfrac{1}{2}((2/3)mR^{2} \times (v/R )^{2})\\\\\\mgD = (5/6)mv^{2}\\\\v = \sqrt{\dfrac{6}{5} gD}[/tex]
Now, apply the second kinematic equation of motion to obtain the time interval for the vertical distance.
[tex]H = ut + \dfrac{1}{2}gt^{2} \\\\H = 0 \times t + \dfrac{1}{2}gt^{2} \\t =\sqrt{\dfrac{2H}{g}}[/tex]
Also, the horizontal distance is,
[tex]L' = v \times t\\\\L'=\sqrt{(6/5)gD} \times \sqrt{2H/g} \\\\L'=2\sqrt{3DH/5[/tex]
Repeat the same procedure for the solid sphere as,
potential energy = kinetic energy
U' = KE'
[tex]mgD=\dfrac{1}{2}mv^{2}+\dfrac{1}{2}I \omega^{2}\\\\\\mgD=\dfrac{1}{2}mv^{2}+\dfrac{1}{2}((2/5)mR^{2} \times (v/R )^{2})\\\\\\mgD = (7/10)mv^{2}\\\\v = \sqrt{\dfrac{10}{7} gD}[/tex]
Also, the horizontal distance covered by solid sphere is,
[tex]L= v \times t\\\\L=\sqrt{(10/7)gD} \times \sqrt{2H/g} \\\\L=2\sqrt{5DH/7[/tex]
Now take the ratio of L' and L as,
[tex]L'/L=2\sqrt{3DH/5} \;/ 2\sqrt{5DH/7}\\\\L'/L= \sqrt{21/5}[/tex]
Thus, the required ratio of L'/L is [tex]\sqrt{21/5}[/tex] .
Learn more about the conservation of energy here:
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