A bowling ball rolls with a speed of 3.72 m/s toward the ball rack, rising h = 0.432 m along the way.
What is the speed of the ball at the top of the incline? Treat the ball as a uniform, solid sphere.
Rotational kinetic energy and inertia?
回答 (1)
2016-11-13 4:33 pm
The ball's initial total and KE = 1/2 mV^2(1 + k) where MOI = kmr^2 is the moment of inertia and k = 2/5 for a solid sphere.
So we have KE - QE = ke = 1/2 mv^2(1 + k) = 1/2 mV^2(1 + k) - mgh. QE = mgh is the work to raise the ball h high.
Which we can reduce to:
v^2(1 + k) = V^2(1 + k) - 2gh; solving for:
v = sqrt(V^2(1 + k) - 2gh)/(1 + k) = sqrt(3.72^2*(1 + 2/5) - 2*9.8*.432)/(1 + 2/5) = 2.36 m/s. ANS.
So we have KE - QE = ke = 1/2 mv^2(1 + k) = 1/2 mV^2(1 + k) - mgh. QE = mgh is the work to raise the ball h high.
Which we can reduce to:
v^2(1 + k) = V^2(1 + k) - 2gh; solving for:
v = sqrt(V^2(1 + k) - 2gh)/(1 + k) = sqrt(3.72^2*(1 + 2/5) - 2*9.8*.432)/(1 + 2/5) = 2.36 m/s. ANS.
收錄日期: 2021-04-22 00:04:19
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https://hk.answers.yahoo.com/question/index?qid=20161113082050AAjVjz7