A ring of mass 3.00kg, inner radius 5.45cm, and outer radius 6.85cm rolls (witho

Question

A ring of mass 3.00kg, inner radius 5.45cm, and outer radius 6.85cm rolls (without slipping) up an inclined plane that makes an angle of ? = 39.1

A ring of mass 3.00kg, inner radius 5.45cm, and outer radius 6.85cm rolls (without slipping) up an inclined plane that makes an angle of ? = 39.1A degree, as shown in the figure below. At the moment the ring is at position x = 2.06m up the plane, its speed is 2.79m/s. The ring continues up the plane for some additional distance and then rolls back down. It does not roll off the top end. How far up the plane does it go?

Explanation / Answer

you will have to find out the moment of inertia[ I ] of the ring........(It has a formula of the form: m(Ra^2+Rb^2) ........just look up for it somewhere)
then the kinetic energy of the ring will be: 0.5*I*((V/r)^2)
potential energy at the topmost position is : m*g*h .....m:mass, g: acceleration due to gravity(9.81), h:vertical height
Using the law of conservation of energy :
kinetic energy + potential energy(at any point) = potential energy (at the topmost position of the ring)..........bcause kinetic energy at the topmost position is zero(ie., the ring would have stopped there)

once you write down the above equation, then you will see that height is the only unknown........solving for which is quite simple.

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