Answer :
Both cylinders will reach the same speed since speed is independent of mass and radius.
Radius of smaller cylinder = R1
Radius of bigger cylinder = R2
their masses = m1/m2 = [tex]\frac{R1}{2R1} ^{2}[/tex]
From law of conservation of energy we have
[tex]mv^{2} /2 + Iw^{2}[/tex] = mgh
[tex]1/2 mv^{2}[/tex] + [tex]m\frac{R}{L}[/tex] × [tex]\frac{x}{y}[/tex] = mgh
3/4 v2 = gh
v= [tex]\sqrt{4/3gh}[/tex]
Since from the above relation, speed is independent of mass and radius, so both cylinders have same speed.
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Neither of the two cylinders moves faster at the bottom. Both cylinders move with the same velocity at the bottom.
By the formula for conservation of energy, for a solid cylinder placed at the top of a ramp and rolling down to the bottom, we have
The kinetic energy of linear motion + rotational kinetic energy = potential energy (1)
Kinetic energy due to linear motion (of the center of mass of cylinder)= [tex]\frac{1}{2} mv^{2}[/tex]
Rotational kinetic energy = [tex]\frac{1}{2}Iw^{2}[/tex]
m = mass of cylinder, v = velocity of the center of mass of the cylinder
I = moment of inertia of solid cylinder = [tex]\frac{1}{2}mR^{2}[/tex] (R = radius of cylinder)
angular velocity w = v/R
g = acceleration due to gravity, h= height of the ramp
From equation (1)
[tex]\frac{1}{2}mv^{2}+ \frac{1}{2}Iw^{2} = mgh[/tex]
⇒[tex]\frac{1}{2}mv^{2}+\frac{1}{2}(\frac{1}{2}mR^{2}* \frac{v^{2} }{R^{2} } ) = mgh[/tex]
⇒[tex]\frac{1}{2}mv^{2}+\frac{1}{4}mv^{2}=mgh[/tex]
⇒[tex]\frac{3}{4}[/tex][tex]mv^{2} =mgh[/tex]
⇒v=√(4gh/3) (cancel m on both sides)
Thus we see the velocity at the bottom of the sloping ramp that is v is independent of the mass or radius of the solid cylinders and only depends on the height of the ramp and acceleration due to gravity.
Thus both cylinders reach the bottom with the same velocity.
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