Question

A hoop, a solid cylinder, a spherical shell, and a solid sphere are placed at rest at the top of an inclined plane. All the objects have the same radius. They are all released at the same time and allowed to roll down the plane. Which object reaches the bottom first?

Answer 1

Answer:

Solid sphere will reach first

Explanation:

When an object is released from the top of inclined plane

Then in that case we can use energy conservation to find the final speed at the bottom of the inclined plane

initial gravitational potential energy = final total kinetic energy

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

now we have

$I = mk^2$

here k = radius of gyration of object

also for pure rolling we have

$v = R\omega$

so now we will have

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(\frac{v^2}{R^2})$

$mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$

$v^2 = \frac{2gh}{1 + \frac{k^2}{R^2}}$

so we will say that more the value of radius of gyration then less velocity of the object at the bottom

So it has less acceleration while moving on inclined plane for object which has more value of k

So it will take more time for the object to reach the bottom which will have more radius of gyration

Now we know that for hoop

$mk^2 = mR^2$

k = R

For spherical shell

$mk^2 = \frac{2}{3}mR^2$

$k = \sqrt{\frac{2}{3}} R$

For solid sphere

$mk^2 = \frac{2}{5}mR^2$

$k = \sqrt{\frac{2}{5}} R$

So maximum value of radius of gyration is for hoop and minimum value is for solid sphere

so solid sphere will reach the bottom at first