Question

A solid sphere rolls without slipping down an incline, starting from rest. At the same time, a box starts from rest at the same altitude and slides down the same incline, with negligible friction. Which object arrives at the bottom first? a) The solid sphere arrives first. b) The box arrives first. c) Both arrive at the same time. d) It is impossible to determine.

Short answer

(a) The solid sphere or (b) The box? Answer: (b) The box arrives first.

Step-by-step solution

Analyze the motion of the box

The box slides down the incline with no friction, so it means its motion follows standard free-fall kinematics, without rotational kinetic energy. The acceleration of the box, which is given by Newton's second law, can be expressed as: \( a_{box} = g \cdot \sin(\theta) \) where \(a_{box}\) is the acceleration of the box, \(g\) is the acceleration due to gravity, and \(\theta\) is the angle of the incline.

Analyze the motion of the solid sphere

As the sphere is rolling down the incline without slipping, we need to consider both its linear and angular components of motion. For the sphere, total kinetic energy is a combination of translational (linear) and rotational (angular) kinetic energy. Therefore, the acceleration of the solid sphere can be expressed as: \( a_{sphere} = \frac{5}{7} g \cdot \sin(\theta) \) This equation represents the acceleration of the solid sphere, considering the no-slipping condition and the specific moment of inertia for a solid sphere which is \(\frac{2}{5} mR^2\).

Compare the accelerations of the box and solid sphere

Now that we have the expressions for the accelerations of both the box and the solid sphere, we can compare them: \( a_{box} = g \cdot \sin(\theta) \) \( a_{sphere} = \frac{5}{7} g \cdot \sin(\theta) \) We see that the acceleration of the box is greater than the acceleration of the solid sphere, so the box will reach the bottom of incline faster than the solid sphere.

Answer

The correct answer is (b) The box arrives first.