Question

A solid uniform sphere and a thin-walled, hollow sphere have the same mass M and radius R. If they roll without slipping up a ramp that is inclined at an angle β above the horizontal and if both have the same vcm before they start up the incline, which object reaches the greater height, or do both of them reach the same height?

Answer 1

Answer:Hollow sphere

Explanation:

Given

same mass for solid and hollow sphere

same $v_{cm}$ before they start up incline

Moment of inertia of solid Sphere

$I_1=\frac{2}{5}Mr^2$

Moment of inertia of hollow sphere

$I_2=\frac{2}{3}Mr^2$

Conserving Energy at bottom and top point for solid sphere

kinetic energy +Rotational Energy=Potential energy

$\frac{1}{2}Mv_{cm}^2+\frac{1}{2}I\omega ^2=mgh_1$

for pure rolling $v_{cm}=\omega r$

$\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\times \frac{2}{5}Mr^2=Mgh_1$

$\frac{7}{10}Mv_{cm}^2=Mgh_1$

$h_1=\frac{7v_{cm}^2}{10g}$

For hollow sphere

$\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\times \frac{2}{3}Mr^2=Mgh_2$

$h_2=\frac{5v_{cm}^2}{6g}$

therefore height gained by hollow sphere is more

Answer 2

The hollow sphere reached the greatest height.

Height reached by each object

The height reached by each object is determined by applying the principle of conservation of energy as show below;

K.E = P.E

¹/₂mv² + ¹/₂Iω² = mgh

where;

• I moment of inertia
• ω is angular speed  = v/r

Height reached by solid sphere

I =  ²/₅mr²

¹/₂mv² + ¹/₂Iω² = mgh

¹/₂mv² + ¹/₂(²/₅mr²)(v/r)² = mgh

¹/₂v² + ¹/₅v² = gh

7v² = 10gh

h = 7v²/10g

h = 0.7(v²/g)

Height reached by hollow sphere

I = ²/₃mr²

¹/₂mv² + ¹/₂Iω² = mgh

¹/₂mv² + ¹/₂(²/₃mr²)(v/r)² = mgh

¹/₂v² + ¹/₃v² = gh

5v² = 6gh

h = 5v²/6g

h = 0.83(v²/g)

Thus, the hollow sphere reached the greatest height.

Learn  more about conservation of energy here: https://brainly.com/question/166559