Question

A frictionless circular ring 1.00 m in diameter is placed flat on the floor, and a 100-g ball is sent around its inside surface with an initial speed of 2.00 m/s. After one full revolution, the speed of the ball measured to be 1.85 m/s. (a) How much energy is converted to internal energy, heating up the ball and the floor, per revolution

Answer 1

`\Delta =2.88* 10^(-2) J`

Step-by-step explanation:

given,

diameter of the ring = 1 m

radius = 0.5 m

mass of the ball = 100 g = 0.1 Kg

initial speed of the ball = 1.85 m/s

converted energy = change in kinetic energy

`\Delta = (1)/(2)m(v_f^2-v_i^2)`

`\Delta = (1)/(2)* 0.1 * (2^2 - 1.85^2)`

`\Delta = (1)/(2)* 0.1 * (2^2 - 1.85^2)`

`\Delta = 0.0288 J`

`\Delta =2.88* 10^(-2) J`