Question

A 20 kg bowling ball with a radius of 28 cmstarts from rest at the top of an incline 2.4 min height.Find the translational speed of the bowlingball after it has rolled to the bottom of theincline. (Assume that the ball is a uniformsolid sphere.)The acceleration of gravity is9.81 m/

Answer 1

v = 5.80 m/s

Step-by-step explanation:

given data

mass = 20 kg

radius = 28 cm

height = 2.4 m

to find out

speed of the bowling

solution

we get here first total energy of the ball on top of incline that is express as

total energy = mgh ................1

total energy = 20 × 9.81 × 2.4

total energy = 470.88 Nm

and here total energy is change in to rotational and kinetic energy

so here

total energy = 0.5 × m × v² + 0.5 × I × ω² ...............2

here v = ωr

470.88 = 0.5 × m × ω²r² + 0.5 ×
`(2)/(5)` × mr² × ω²

470.88 = 0.5 × m × ω²r² ( 1+
`(2)/(5)` )

470.88 =
`(7)/(10)` × 20 × ω²r²

470.88 = 14 × v²

v² = 33.6342

v = 5.80 m/s