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Physics

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You roll a 5 kg bowling ball (R=20cm) with an initial speed of 10 m/s up a 30 degree ramp that is 12 m long. Will the ball reach the top of the ramp? (support your answer)

  • Physics -

    What is the translational (1/2 mv^2) PLUS the initial rotational energy (1/2 I w^2, where w=vr). Will that equal mgh?

  • Physics -

    Thanks bobpursley! I was somewhere close to that idea. I had mgh=1/2mv^2 + 1/5mv^2. I let h=12sin30. I was solving for velocity for some reason, doh!

  • Physics -

    so I have mgh = 300.
    i have my rotational energy = 350.
    so it will reach the top, correct?

  • Physics -

    Try to obtain the general solution of the problem, i.e.,
    KE=PE
    KE(tr) + KE(rot) =PE

    0.7m•v²/2 = m•g•h
    h=0.7•v²/g=70/9.8=7.14 m.

    h(real)=s•sinα=12•0.5=6 m
    Since 7,14>6, the ball‘ll
    rich the top of the ramp.

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