Question

Please submit the answer to this problem on a separate piece of paper, in submission box. A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (ho = 20.0 cm). a) Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain b) Again, without doing any calculations, decide which object would get to the bottom first. Hint: which one has greater translational speed? Think CoE! C) Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

Answer 1

a) The moment of inertia of ring is maximum and that of solid sphere is minimum.

When the objects rolls down without slipping, it is the rotational kinetic energy that makes the difference. Higher the moment of inertia greater will be the gain in rotational kinetic energy and less translational kinetic energy.

And the body who has least translational kinetic energy will reach the bottom with least velocity,i.e; the sphere reaches the bottom with highest spinning and the ring with slowest spinning.

b) By the same concept as explained earlier in part a, that the sphere's velocity is maximum at the bottom and so it is fastest to reach the bottom and the ring will reach at last.

c) expression of translation kinetic energy = $K_{t}=\frac{1}{2}mv^{2}$

expression of rotational kinetic energy = $K_{r}=\frac{1}{2}I\omega^{2}=\frac{1}{2}\frac{I}{R^{2}}v^{2}$ [since in pure rolling, angular speed = linear speed/radius on fixed surface]

total kinetic energy = $K_{t}+K_{r}=\frac{1}{2}mv^{2}\left ( 1+\frac{I}{mR^{2}} \right )$

fraction of translation kinetic energy = $\frac{K_{t}}{K_{total}}=\frac{\frac{1}{2}mv^{2}}{\frac{1}{2}mv^{2}\left ( 1+\frac{I}{mR^{2}} \right )}=\frac{1}{1+\frac{I}{mR^{2}}}$

Given:

h = 20 cm = 0.2 m

m = 2 kg

R = 3 cm = 0.03 m

As the mechanical energy of the body is always conserved, the total kinetic energy at the bottom is equal to the potential energy at highest when the the body starts its motion with zero initial velocity.

Solving the problem for sphere:

Therefore, $\frac{K_{t}}{K_{total}}=\frac{1}{1+\frac{I}{mR^{2}}}$

$\Rightarrow K_{t}=\frac{1}{1+\frac{I}{mR^{2}}}*K_{total}$

$\Rightarrow \frac{1}{2}mv^{2}=\frac{1}{1+\frac{\frac{2}{5}mR^{2}}{mR^{2}}}*mgh$

$\Rightarrow \frac{1}{2}v^{2}=\frac{1}{1+\frac{2}{5}}*gh$

$\Rightarrow v^{2}=\frac{10}{7}*gh$

$\Rightarrow v=\sqrt{\frac{10gh}{7}}=\sqrt{\frac{10*9.81*0.2}{7}}=1.67m/s$ [answer]

Rotationa speed = angular speed = $\frac{v}{R}=\frac{1.67}{0.03}=55.67rad/s$ [answer]