Physics, linear speed?
After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s . To reach the rack, the ball rolls up a ramp that rises through a vertical distance of 0.53m
What is the linear speed of the ball when it reaches the top of the ramp? in m/s
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The total kinetic energy of the ball when it reaches the ramp is
KE1 = 0.5*m*v1² + 0.5*I*ω1²
ω1 = v1/r and I for a solid sphere is (2/5)*m*r²
KE1 = (0.5 + 0.2)*m*v1²
KE1 = 0.7*m*v1²
On going up the ramp, it loses energy = m*g*h, so the kinetic energy at the top is
KE2 = 0.7*m*v1² - m*g*h
KE2 is also equal to 0.7*m*v2²
0.7*m*v2² = 0.7*m*v1² - m*g*h
v2 = √[v1² - g*h/0.7]