Question

A spherically symmetric object, with radius R=0.60 m and mass M=2.7 kg, rolls without slipping across a horizontal floor, with velocity V=2.5 m/s. It then rolls up an incline with an angle of inclination θ=31 ∘ and comes to rest a distance d=3.8 m up the incline, before reversing direction and rolling back down. Find the moment of inertia of this object about an axis through its center of mass.

Answer 1

The moment of inertia of an object that rolls without slipping down an incline can be calculated using the conservation of energy equation and the relation between angular velocity and radius.

The moment of inertia of an object that rolls without slipping down an incline can be expressed as a multiple of MR², where M is the mass of the object and R is its radius. To find the moment of inertia, we need to use the conservation of energy equation: mgh = (1/2)mv² + (1/2)Iω². Since the object starts from rest at the top of the incline, the initial kinetic energy is zero. Thus, the equation becomes mgh = (1/2)mv² + (1/2)Iω². We know the height of the incline (h = 2.00 m), the final velocity (v = 6.00 m/s), and we can express the angular velocity ω in terms of the radius using the relation ω = v/R. Plugging in these values, we can solve for I.

A physical property called the moment of inertia quantifies how easily a body may rotate along a specific axis. It is the resistance of an item to translational motion; it is the rotational analogue of mass. Matter possesses the quality of inertia, which prevents it from changing its velocity.