A solid uniform sphere and a uniform spherical shell, both having the same mass

Question

A solid uniform sphere and a uniform spherical shell, both having the same mass and radius, roll without slipping down a hill that rises at an angle (theda) above the horizontal. Both spheres start from rest at the same vertical height h.


How fast is each sphere moving when it reaches the bottom of the hill?

(Make sure that you know how to derive the answer, an algebraic expression, step by step starting from basic equations!)

Vsolid=

Vhollow=

Which will reach the bottom first?

Please help me understand!!!!!! :/

Explanation / Answer

Loss of potential energy =Gain in kinetic enrgy
U = K
mgh = 0.5I2 + 0.5mv2
2mgh = I2 + mv2
I1 =(2/5) mr2 for solid sphere
and I2 =(2/3)mr2 for hollow sphere
Alo 2 r2= v2
2mgh = =(2/5)mr2 2 + mv2
2gh =v2 (2/5 + 1) =v2  (7/5 for solid sphere
And 2gh =v2  (2/3+ 1) =v2 (5/3 for shell

Hence, for solid sphere,
v12 = (10/7)gh; v1 =[(10/7)gh]
For hollow sphere, v22 =(6/5)gh; v2 =[(6/5)gh]

So, the hollow sphere has the smaller moment of inertia => greater angular acceleration so it will reach the bottom of the slope first.

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