Answer:
vcyl / vsph = 1.05
Explanation:
- The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
- The traslational part can be written as follows:
[tex]K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)[/tex]
- The rotational part can be expressed as follows:
[tex]K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)[/tex]
- where I = moment of Inertia regarding the axis of rotation.
- ω = angular speed of the rotating object.
- If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:
[tex]v = \omega * R (3)[/tex]
- For a solid cylinder, I = M*R²/2 (4)
- Replacing (3) and (4) in (2), we get:
[tex]K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)[/tex]
- Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:
[tex]K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)[/tex]
- Repeating the same steps for the spherical shell:
[tex]I_{sph} = \frac{2}{3} * M* R^{2} (7)[/tex]
[tex]K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)[/tex]
[tex]K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)[/tex]
- Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
- And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
- Rearranging, and taking square roots on both sides, we get:
[tex]\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)[/tex]
- This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.
