Respuesta :
Answer:
[tex]1.2:1[/tex]
Explanation:
For the marble :
[tex]h[/tex] = height from which the marble is released
[tex]v_{m}[/tex] = speed of center of mass of marble at the bottom
[tex]w[/tex] = angular speed of marble = [tex]\frac{v_{m}}{r}[/tex]
[tex]m[/tex] = mass of the marble
[tex]r[/tex] = radius of the marble
Moment of inertia of marble (solid sphere) is given as
[tex]I = (0.4) m r^{2}[/tex]
Using conservation of energy between Top and bottom
Kinetic energy at bottom + rotational kinetic energy at bottom = potential energy at top
[tex](0.5)mv_{m}^{2} + (0.5)Iw^{2}= mgh[/tex]
[tex]mv_{m}^{2} + (0.4) mr^{2}\left ( \frac{v_{m}}{r} \right )^{2}= 2mgh[/tex]
[tex]mv_{m}^{2} + (0.4) mv_{m}^{2}= 2mgh[/tex]
[tex](1.4)v_{m}^{2} = 2gh[/tex] Eq-1
For Cube :
[tex]h[/tex] = height from which the cube is released
[tex]v_{c}[/tex] = speed of center of mass of cube at the bottom
[tex]m[/tex] = mass of the cube
Using conservation of energy between Top and bottom
Kinetic energy at bottom = potential energy at top
[tex](0.5)mv_{c}^{2} = mgh[/tex]
[tex]v_{c}^{2} = 2gh[/tex] Eq-2
Using Eq-1 and Eq-2
[tex]v_{c}^{2} = (1.4)v_{m}^{2}[/tex]
Taking square-root both side
[tex]\sqrt{v_{c}^{2}} = \sqrt{(1.4) v_{m}^{2}}[/tex]
[tex]\sqrt{v_{c}^{2}} = \sqrt{(1.4) v_{m}^{2}}\\\\v_{c} = 1.2 v_{m}[/tex]
[tex]\frac{v_{c}}{v_{m}} = 1.2[/tex]
