A solid spherical ball (M = 14 kg; R = 30 cm) is rolling along a level surface at 25 m/s when it encounters a ramp that has a length of 15 m and is declined at an angle of 30° to the horizontal. The ball continues to roll without slipping down the ramp. Determine the magnitude of the velocity of the ball at the bottom of the ramp.
- NCSLv 74 months agoFavourite answer
For a solid sphere, the moment of inertia about an axis through its center is
I = (2/5)mR²
"roll without slipping" means that
ω = v/R
so the total KE is
KE = ½mv² + ½Iω² = ½mv² + ½*[(2/5)mR²]*(v/R)² = 0.7mv²
final KE = initial KE + mgh
mass m cancels
0.7*V² = 0.7*(25m/s)² + 9.8m/s² * 15m*sin30º = 511 m²/s²
final V = 27 m/s
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- oubaasLv 74 months ago
0,7*m*Vo*2 + m*g*h = 0,7*m*Vf^2
Vf = √ 0,7Vo^2/0,7+g*h/0,7 = √25^2+9,807*15*0.5/0.7 = 27.0 m/sec