A solid sphere with radius r=2m rolls smoothly from rest from a ramp at height h=10m that is the vertical distance between the COM of the sphere and the bottom of the ramp . When the sphere reached the bottom of the ramp, the sphere introduces a circular path with R. When the sphere reaches the the top of the circular path with the speed of 3m/s determine the radius of the circle (R) sphere.COM=(2/5)mr^2. assume that the sphere only makes smooth rolling] Select one: a. 4.8 m b. 4.1 m c. 5.7 m d. 4.7 m e. 3.5 m

To solve this problem, we need to use the principles of conservation of energy and the relationship between the center of mass (COM) and the radius of the sphere.<br /><br />Given information:<br />- Radius of the sphere, $r = 2m$<br />- Height of the ramp, $h = 10m$<br />- Speed of the sphere at the top of the circular path, $v = 3m/s$<br /><br />Step 1: Calculate the potential energy at the top of the ramp.<br />Potential energy at the top of the ramp = $mgh$<br />where $m$ is the mass of the sphere and $g$ is the acceleration due to gravity.<br /><br />Step 2: Calculate the kinetic energy at the bottom of the ramp.<br />Since the sphere rolls smoothly, the kinetic energy at the bottom of the ramp is the sum of the translational kinetic energy and the rotational kinetic energy.<br />Kinetic energy at the bottom of the ramp = $\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$<br />where $I$ is the moment of inertia of the sphere and $\omega$ is the angular velocity.<br /><br />Step 3: Use the conservation of energy principle to find the radius of the circular path.<br />Potential energy at the top of the ramp = Kinetic energy at the bottom of the ramp<br />$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$<br /><br />Step 4: Substitute the values and solve for the radius of the circular path.<br />Given that the COM of the sphere is $(2/5)mr^2$, we can use this relationship to find the moment of inertia of the sphere.<br />$I = \frac{2}{5}mr^2$<br /><br />Substituting the values, we get:<br />$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2$<br />$10m \cdot m \cdot 9.8m/s^2 = \frac{1}{2}m(3m/s)^2 + \frac{1}{2}\left(\frac{2}{5}m(2m)^2\right)\left(\frac{3m/s}{2m}\right)^2$<br />$98m = 4.5m + 1.8m$<br />$98m = 6.3m$<br /><br />Solving for the radius of the circular path, we get:<br />$R = \frac{v}{\sqrt{\frac{2gh}{m}}}$<br />$R = \frac{3m/s}{\sqrt{\frac{2 \cdot 9.8m/s^2 \cdot 10m}{m}}}$<br />$R = \frac{3m/s}{\sqrt{196m/s^2}}$<br />$R = \frac{3m/s}{14m/s}$<br />$R = 0.214m$<br /><br />Therefore, the correct answer is:<br />b. 4.1 m