A solid sphere of mass 4.0 kg and radius 0.12 m starts from rest at the top of a ramp inclined 15°, and rolls to the bottom. The upper end of the ramp is 1.2 m higher than the lower end. What is the linear speed of the sphere when it reaches the bottom of the ramp

As per mechanical energy conservation law we can say that initial total gravitational potential energy of the sphere is equal to final kinetic energy of rolling

so we will say

[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/tex]

now for pure rolling condition we know that

[tex]v = R\omega[/tex]

so we will have

[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2})[/tex]

[tex]mgh = \frac{7}{10}mv^2[/tex]

now we will have

[tex]v^2 = \sqrt{\frac{10}{7}gh}[/tex]

[tex]v^2 = \sqrt{\frac{10}{7}(9.8)(1.2)}[/tex]

[tex]v = 4.1 m/s[/tex]

okpalawalter8 okpalawalter8 · Answer 2 · 2022-03-28T11:57:55+02:00

The Linear speed of the sphere is mathematically given as

v = 4.1 m/s

What is the linear speed of the sphere when it reaches the bottom of the ramp?

Question Parameter(s):

A solid sphere of mass 4.0 kg and radius of 0.12 m starts from rest at the top of a ramp inclined 15°

Generally, the equation for the conservation of energy is mathematically given as

[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}Iw^2[/tex]

Therefore

[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2})\\\\mgh = \frac{7}{10}mv^2[/tex]

In conclusion, the Speed is

[tex]v^2 = \sqrt{\frac{10}{7}(9.8)(1.2)}[/tex]

v = 4.1 m/s