a solid sphere with a uniform surface density is rolling over a plane. its rotat
ID: 1411300 • Letter: A
Question
a solid sphere with a uniform surface density is rolling over a plane. its rotational kinetic energy is 80j. what is the kinetic energy of the linear motion ( translational kinetic energy) and the total kinetic energy? a solid sphere with a uniform surface density is rolling over a plane. its rotational kinetic energy is 80j. what is the kinetic energy of the linear motion ( translational kinetic energy) and the total kinetic energy? a solid sphere with a uniform surface density is rolling over a plane. its rotational kinetic energy is 80j. what is the kinetic energy of the linear motion ( translational kinetic energy) and the total kinetic energy?Explanation / Answer
Hi,
In this case we have to consider the way to calculate the kinetic energy of an object that is rotating and travelling along a certain trajectory, as well as the shape of the object.
The following relations are well known and they are neccesary to solve this problem:
K = Kr + Kc ; where Kr is the rotational kinetic energy and Kc is the translational kinetic energy
Kc = (1/2) mv2 ; where m is the mass of the object and v is the translational velocity
Kr = (1/2) Iw2 ; where I is the moment of inertia of the object and w is the rotational speed
I = (2/5) mR2 (moment of inertia of a solid sphere); where R is the radius of the sphere
w = v/R ; the relation between the linear and the rotational speed
So:
Kr = (1/2) Iw2 = (1/2) [(2/5) mR2] (v/R)2 = (1/5) mv2
Kc = (1/2) (5 Kr) ::::::: Kc = (5/2) (80 J) = 200 J (this is the kinetic energy of the linear movement)
K = Kc + Kr = 200 J + 80 J = 280 J (this is the total kinetic energy of the object)
I hope it helps.
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