Respuesta :
Explanation:
For each object, the initial potential energy is converted to rotational energy and translational energy:
PE = RE + KE
mgh = ½ Iω² + ½ mv²
For the marble (a solid sphere), I = ⅖ mr².
For the basketball (a hollow sphere), I = ⅔ mr².
For the manhole cover (a solid cylinder), I = ½ mr².
For the wedding ring (a hollow cylinder), I = mr².
If we say k is the coefficient in each case:
mgh = ½ (kmr²) ω² + ½ mv²
For rolling without slipping, ωr = v:
mgh = ½ kmv² + ½ mv²
gh = ½ kv² + ½ v²
2gh = (k + 1) v²
v² = 2gh / (k + 1)
The smaller the value of k, the higher the velocity. Therefore:
marble > manhole cover > basketball > wedding ring
The arrangement of the given objects in order of their increasing speed down the ramp is wedding ring > large marble > manhole cover > basketball.
Apply the principle of conservation of energy;
[tex]\frac{1}{2} mv^2 \ + \ \frac{1}{2} I \omega ^2 = mgh[/tex]
where;
- I is moment of inertia
The moment of inertia for the marble (a solid sphere);
[tex]I = \frac{2}{5} mr^2[/tex]
The moment of inertia for the basketball (a hollow sphere);
[tex]I = \frac{2}{3} mr^2[/tex]
The moment of inertia for the manhole cover (a solid cylinder);
[tex]I = \frac{1}{2} mr^2[/tex]
The moment of inertia for the wedding ring (a hollow cylinder);
[tex]I = mr^2[/tex]
[tex]\frac{1}{2} mv^2 \ + \ \frac{1}{2} I (\frac{v}{r} )^2 = mgh\\\\\frac{1}{2} mv^2 \ + \ \frac{1}{2r^2} Iv^2 = mgh\\\\v^2 (m + \frac{I}{r^2} )= 2mgh\\\\v^2 = \frac{2mgh}{m + \frac{I}{r^2} } \\\\v = \sqrt{\frac{2mgh}{m + \frac{I}{r^2} } }[/tex]
The smaller the value of moment of inertia, the greater the value the speed of the object.
The arrangement of the given objects in order of their increasing speed down the ramp;
wedding ring > large marble > manhole cover > basketball
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