A hoop of radius RH and mass mH and a solid cylinder of radius Rc and mass mc are released simultaneously at the top of a plane ramp of length L inclined at angle θ above horizontal. Which reaches the bottom first, and what is the speed of each there?
The Correct Answer and Explanation is :
When two objects, such as a hoop and a solid cylinder, are released simultaneously from the top of a ramp, their motion will depend on their moments of inertia and how they roll down the incline. Both objects will undergo rolling motion, and to determine which reaches the bottom first and their speeds, we must analyze their energy equations.
1. Moment of Inertia and Rolling Motion:
The key difference between the hoop and the solid cylinder lies in their moments of inertia:
- Moment of inertia for the hoop: ( I_H = m_H R_H^2 )
- Moment of inertia for the solid cylinder: ( I_C = \frac{1}{2} m_C R_C^2 )
Where ( R_H ) and ( R_C ) are the radii of the hoop and the solid cylinder, and ( m_H ) and ( m_C ) are their respective masses.
Both objects start from rest, so they convert potential energy to both translational and rotational kinetic energy as they roll down the ramp. The total mechanical energy at the top of the ramp is purely potential energy:
[
PE = mgh
]
where ( h ) is the height of the ramp (related to the length ( L ) and angle ( \theta ) by ( h = L \sin\theta )).
As they roll, the energy splits into translational and rotational forms:
[
E_{\text{total}} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2
]
Since the objects roll without slipping, ( v = \omega R ), so the rotational kinetic energy can be rewritten as:
[
E_{\text{total}} = \frac{1}{2} m v^2 + \frac{1}{2} \frac{I}{R^2} v^2 = \left( \frac{1}{2} m + \frac{1}{2} \frac{I}{R^2} \right) v^2
]
For each object, the total energy is equal to the initial potential energy ( mgh ):
[
mgh = \left( \frac{1}{2} m + \frac{1}{2} \frac{I}{R^2} \right) v^2
]
Solving for the velocity ( v ) at the bottom, we get:
[
v = \sqrt{\frac{2mgh}{m + \frac{I}{R^2}}}
]
2. Comparing the Two Objects:
For the hoop:
[
v_H = \sqrt{\frac{2 m_H g h}{m_H + \frac{m_H R_H^2}{R_H^2}}} = \sqrt{\frac{2 g h}{2}} = \sqrt{g h}
]
For the solid cylinder:
[
v_C = \sqrt{\frac{2 m_C g h}{m_C + \frac{1}{2} m_C}} = \sqrt{\frac{2 g h}{\frac{3}{2}}} = \sqrt{\frac{4}{3} g h}
]
Since ( \sqrt{\frac{4}{3}} > 1 ), the solid cylinder will reach the bottom of the ramp first.
3. Conclusion:
- Which reaches the bottom first? The solid cylinder reaches the bottom first because its moment of inertia is smaller, meaning it accelerates faster.
- Speeds at the bottom:
- Speed of the hoop: ( v_H = \sqrt{g h} )
- Speed of the solid cylinder: ( v_C = \sqrt{\frac{4}{3} g h} )
Thus, the solid cylinder is both faster and reaches the bottom first.