A group of students perform the same "Conservation of Mechanical Energy" experiment that you performed in lab by allowing a solid sphere and then a solid cylinder to roll down the ramp. Both objects were released from the same position at the top of the ramp. If the speed vsphere of the solid sphere at the bottom of the ramp was 2.25 m/s, what would be the speed vcylinder at the bottom of the ramp

2.17 m/s

Explanation:

We can solve this problem by using the conservation of mechanical energy.

In fact, for each of the two objects, the initial gravitational potential energy when it is at the top of the ramp is entirely converted into translational kinetic energy + rotational kinetic energy. So we can write for each object

(1)

where

is the gravitational potential energy, with

m = mass of the object

is the acceleration due to gravity

h is the initial height of the object

is the translational kinetic energy, where

v is the final speed of the object

is the rotational kinetic energy, where

I is the moment of inertia of the object

is the final angular velocity

We know that for the solid sphere, the speed at the bottom of the ramp is

v = 2.25 m/s

Moreover for a solid sphere,

where r is the radius of the sphere. And using the relationship between angular speed and linear speed, , we can the rotational energy as

So eq.(1) for the solid sphere becomes:

And solving for h, we find the height of the ramp:

Now we analyze the solid cylinder; its moment of inertia is

So its rotational energy can be written as

So eq.(1) for the cylinder is

And since we now know the value of h, we can find the final speed: