Answer:
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
[tex]mgh=\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/tex] (1)
where
[tex]mgh[/tex] is the gravitational potential energy, with
m = mass of the object
[tex]g=9.8m/s^2[/tex] is the acceleration due to gravity
h is the initial height of the object
[tex]\frac{1}{2}mv^2[/tex] is the translational kinetic energy, where
v is the final speed of the object
[tex]\frac{1}{2}I\omega^2[/tex] is the rotational kinetic energy, where
I is the moment of inertia of the object
[tex]\omega[/tex] 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,
[tex]I=\frac{2}{5}mr^2[/tex]
where r is the radius of the sphere. And using the relationship between angular speed and linear speed, [tex]v=\omega r[/tex], we can the rotational energy as
[tex]\frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v^2}{r^2})=\frac{1}{5}mv^2[/tex]
So eq.(1) for the solid sphere becomes:
[tex]mgh=\frac{1}{2}mv^2 + \frac{1}{5}mv^2\\mgh=\frac{7}{10}mv^2[/tex]
And solving for h, we find the height of the ramp:
[tex]h=\frac{7v^2}{10g}=\frac{7(2.25)^2}{10(9.8)}=0.36 m[/tex]
Now we analyze the solid cylinder; its moment of inertia is
[tex]I=\frac{1}{2}mr^2[/tex]
So its rotational energy can be written as
[tex]\frac{1}{2}I\omega^2 = \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v^2}{r^2})=\frac{1}{4}mv^2[/tex]
So eq.(1) for the cylinder is
[tex]mgh=\frac{1}{2}mv^2+\frac{1}{4}mv^2=\frac{3}{4}mv^2[/tex]
And since we now know the value of h, we can find the final speed:
[tex]v=\sqrt{\frac{4gh}{3}}=\sqrt{\frac{4(9.8)(0.36)}{3}}=2.17 m/s[/tex]
