Question

Derive an expression for the total kinetic energy of a body rolling on a horizontal
plane without slipping.​

Answer 1

SOLUTION.

Let the mass of the body be m

the velocity of the body is v

the angular velocity of the body is w

TRANSLATIONAL KINETIC ENERGY

1/2mv²

Rotational kinetic energy

1/2Iw²

where I is the moment of inertia of the body which can be written as Kmr²

where r is the radius

K is a constant

Total Kinetic energy

1/2mv² + 1/2Iw²

(1/2)×m×[v² + kw²r²]

(m/2)(v² + v²k)...............v = wr ( without slliping )

(m/2)(v²)(1 + k)

or

(m/2)(w²r²)(1 + k)

#answerwithquality

#BAL